NCERT Solutions for Class 11 Chemistry Chapter 2 structure of the atom is a critical study material that will boost your preparations for class 11 chemistry examination and entrance examinations. These solutions provide you with comprehensive answers to the question given in the textbook.
Class 11 NCERT Solutions for chemistry chapter 2 Structure of the atom
NCERT Solutions Class 11 Chemistry Chapter 2 help students learn and practice atomic structure questions and different concepts related to it. Solutions for NCERT class 11 Chemistry chapter 2 structure of atom includes solutions for all the questions provided in class 11 chemistry NCERT textbook along with answers to extra questions and important question of previous year question papers.
Students can practice various questions about atoms, atomic structure, protons, neutrons electrons, etc. These solutions for class 11 Chemistry chapter 2 provides different examples followed by the solved exercise, exemplary problems and worksheets for the student’s benefits. These study materials include both objective, long and short type of questions related to this topic.
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Subtopics Of Class 11 Chemistry Chapter 2 Structure Of The Atom

Subatomic Particles
 Discovery Of Electron
 Charge To Mass Ratio Of Electron
 Charge On The Electron
 Discovery Of Protons And Neutrons

Atomic Models
 Thomson Model Of Atom
 Rutherford’s Nuclear Model Of Atom
 Atomic Number And Mass Number
 Isobars And Isotopes
 Drawbacks Of Rutherford Model

Developments Leading To The Bohr’s Model Of Atom
 Wave Nature Of Electromagnetic Radiation
 Particle Nature Of Electromagnetic Radiation: Planck’s Quantum Theory
 Evidence For The Quantized* Electronic Energy Levels: Atomic Spectra

Bohr’s Model For Hydrogen Atom
 Explanation Of Line Spectrum Of Hydrogen
 Limitations Of Bohr’s Model

Towards Quantum Mechanical Model Of The Atom
 Dual Behaviour Of Matter
 Heisenberg’s Uncertainty Principle

Quantum Mechanical Model Of Atom
 Orbitals And Quantum Numbers
 Shapes Of Atomic Orbitals
 Energies Of Orbitals
 Filling Of Orbitals In Atom
 Electronic Configuration Of Atoms
 Stability Of Completely Filled And Half Filled Subshells.
Class 11 Chemistry Chapter 2 Structure Of The Atom Important questions
Q.1. (a)Figure the quantity of electrons which will together measure one gram.
(b)Compute the mass and charge of one mole of electrons.
Ans:
(a)Mass of one electron =\(9.10939\times 10^{31}\: kg\) \(∴\) Quantity of electrons that measure \(9.10939\times 10^{31}\: kg\)=1
\(∴\) Quantity of electrons that will measure 1 g (\(1\times 10^{3}kg\)) \(=\frac{1}{9.10939\times 10^{31}kg}\times (1\times 10^{3}kg)\\ \\ =0.1098\times 10^{3+31}\\ \\ =0.1098\times 10^{28}\\ \\ =1.098\times 10^{27}\)(b)Mass of one electron =\(9.10939\times 10^{31}\: kg\)
Mass of one mole of electron =\((6.022\times 10^{23})\times (9.10939\times 10^{31}kg)\\ \\ =5.48\times 10^{7}kg\)
Charge on one electron =\(1.6022\times 10^{19}coulomb\)
Charge on one mole of electron =\((1.6022\times 10^{19}C)(6.022\times 10^{23})=9.65\times 10^{4}C\)
Q.2. (I)Figure the aggregate number of electrons present in one mole of methane.
(II)Discover (a) the aggregate number and (b) the aggregate mass of neutrons in 7 mg of 14C.
(Assume that mass of a neutron\(= 1.675 × 10^{27} kg\)).
(III)Discover (a) the aggregate number and (b) the aggregate mass of protons in 34 mg of \(NH_{3}\) at STP.
Will the appropriate response change if the temperature and weight are changed?
Ans:
(I)Number of electrons present in 1 molecule of methane (\(CH_{4}\))
\({1(6)+4(1)}=10\)Number of electrons present in 1 mole i.e., \(6.023\times 10^{23}\) molecules of methane
\(=6.022\times 10^{23}\times 10=6.022\times 10^{24}\)(II)(a) Number of atoms of \(14_{C}\) in 1 mole=\(6.023 \times 10^{23}\)
Since 1 atom of \(14_{C}\) contains (14 – 6) i.e., 8 neutrons, the number of neutrons in 14 g of \(14_{C}\) is \((6.023 \times 10^{23})\times8\).Or, 14 g of \(14_{C}\) contains \((6.023 \times 10^{23}\times8)\)neutrons.
Number of neutrons in 7 mg
\(\frac{ 6.022 \times 10^{23}\times 8\times 7\, mg}{1400\, mg}\\ \\ =2.4092\times 10^{21}\)
(b)The aggregate mass of one neutron= \(1.67493 \times 10^{–27}\, kg\)
The aggregate mass of total neutrons in 7 g of \(14_{C}\) \(= (2.4092 \times 10^{21}) (1.67493 \times 10^{–27}\, kg) \\ \\ =4.0352\times 10^{6}\, kg\)
(III)(a)1 mole of \(NH_{3}={1(14)+3(1)}\, g\: of\: NH_{3}\\ \\ =17\, g\: of\: NH_{3}\\ \\ =6.022\times 10^{23}\: molecules \: of\: NH_{3}\)
Total number of protons present in 1 molecule of \(NH_{3}\) \(={1(7)+3(1)}\\ \\ =10\)
Number of protons in \(6.023\times 10^{23}\: molecules\: of\: NH_{3}\) \(=(6.023\times 10^{23})(10)\\ \\ =6.023\times 10^{24}\\ \\ \Rightarrow 17 \, g \: of\: NH_{3}\) contains \((6.023\times 10^{24})\) protons. Number of protons in 34 mg of \(NH_{3}\) \(\frac{6.022\times 10^{24}\times 34\, mg}{17000\, mg}\\ \\ =1.2046\times 10^{22}\)
(b)Mass of one proton = \(1.67493 \times 10^{–27}\, kg\)
Total mass of protons in 34 mg of \(NH_{3}\) \(=(1.67493\times 10^{27}\, kg)(1.2046\times 10^{22} )\\ \\ =2.0176\times 10^{5}\, kg\)
The quantity of protons, electrons, and neutrons in a particle is free of temperature what’s more, weight conditions. Subsequently, the acquired qualities will stay unaltered if the temperature and weight is changed.
Q.3.What number of neutrons and protons are there in the following nuclei ?
\(_{6}^{13}\textrm{C}\: ,\: _{8}^{16}\textrm{O}\: ,\: _{12}^{24}\textrm{Mg}\: ,\: _{16}^{56}\textrm{Fe}\: ,\: _{38}^{88}\textrm{Sr}\)
Ans:
\(_{6}^{13}\textrm{C}\):Atomic mass = 13
Atomic number = Number of protons = 6
Number of neutrons = (Atomic mass) – (Atomic number) = 13 – 6 = 7
\(_{8}^{16}\textrm{O} \):Atomic mass = 16
Atomic number = 8
Number of protons = 8
Number of neutrons = (Atomic mass) – (Atomic number) = 16 – 8 = 8
\( _{12}^{24}\textrm{Mg} \):Atomic mass = 24
Atomic number = Number of protons = 12
Number of neutrons = (Atomic mass) – (Atomic number) = 24 – 12 = 12
\( _{16}^{56}\textrm{Fe} \):Atomic mass = 56
Atomic number = Number of protons = 26
Number of neutrons = (Atomic mass) – (Atomic number) = 56 – 26 = 30
\( _{38}^{88}\textrm{Sr} \):Atomic mass = 88
Atomic number = Number of protons = 38
Number of neutrons = (Atomic mass) – (Atomic number) = 88 – 38 = 50
Q.4. Compose the total image for the atom with the given nuclear number (Z) and Atomic mass (A)
(I)Z = 17, A = 35
(II)Z = 92, A = 233
(III)Z = 4, A = 9
Ans:
(I)\(_{17}^{35}\textrm{C}\)
(II)\(_{92}^{233}\textrm{U}\)
(III)\(_{4}^{9}\textrm{Be}\)
Q.5.Yellow light radiated from a sodium light has a wavelength (\(\lambda\)) of 580 nm. Ascertain the frequency (\(\nu\)) and wave number (\(\bar{\nu }\) ) of the yellow light.
Ans: From the expression,
\(\lambda =\frac{c}{\nu }\)We get,
\(\nu =\frac{c}{ \lambda }\) ……….(1)Where , \(\nu\)= frequency of yellow light
c = velocity of light in vacuum = \(3\times 10^{8}\, m/s\) \(\lambda\)= = wavelength of yellow light = 580 nm = \(580\times 10^{9}\, m/s\)
Substituting the values in expression (1):
\(\nu =\frac{3\times 10^{8}}{580\times 10^{9}}=5.17\times 10^{14}\, s^{1}\)Thus, frequency of yellow light emitted from the sodium lamp =
\(5.17\times 10^{14}\, s^{1}\)Wave number of yellow light \(\bar{\nu }=\frac{1}{\lambda }\) \(=\frac{1}{580\times 10^{9}}=1.72\times 10^{6}\, m^{1}\)
Q.6.Find energy of each of the photons which
(I)Correspond to light of frequency \(3\times 10^{15}\, Hz\).
(II)Have wavelength of 0.50 armstrong.
Ans:
(I)Energy (E) of a photon is given by the expression,
E= \(h\nu\)
Where, h = Planck’s constant = \(6.626\times 10^{34}\, Js\) \(\nu\) = frequency of light = \(3\times 10^{15}\)Hz
Substituting the values in the given expression of E:
\(E=(6.626\times 10^{34})(3\times 10^{15})\\ \\ E=1.988\times 10^{18}\, J\)(II) Energy (E) of a photon having wavelength (\(\lambda\)) is given by the expression,
E= \(hc\nu\)
h = Planck’s constant = \(6.626\times 10^{34}Js\)
c = velocity of light in vacuum = \(3\times 10^{8}\, m/s\)
Substituting the values in the given expression of E:
\(E=\frac{(6.626\times 10^{34})(3\times 10^{8})}{0.50\times 10^{10}}=3.976\times 10^{15}J\\ \\ ∴ E=3.98\times 10^{15}J\)
Q.7.Calculate the wavelength, frequency and wave number of a light wave whose period is \(2.0 \times 10^{–10} \, s\)
Ans: Frequency (\(\nu\)) of light = \(\frac{1}{Period}\) \(\frac{1}{Period}\\ \\ =\frac{1}{2.0\times 10^{10}\, s}\\ \\ =5.0\times 10^{9}\, s^{1 }\)
Wavelength(\(\lambda\)) of light=\(c\nu\)
Where,
c = velocity of light in vacuum =\(3\times 10^{8}\, m/s\)
Substituting the value in the given expression of \(\lambda\):
\(\lambda =\frac{3\times 10^{8}}{5.0\times 10^{9}}=6.0\times 10^{2}m\)Wave number \((\bar{\nu })\) of light = \(\frac{1}{\lambda }=\frac{1}{6.0\times 10^{2}}=1.66\times 10^{1}\, m^{1}=16.66\, m\)
Q.8.What is the quantity of photons of light with a wavelength of 4000 pm that give 1 J of energy?
Ans: Energy (E) of a photon = \(h\nu\)
Energy (\(E_{n}\)) of ‘n’ photons =\(nh\nu\) \(\Rightarrow n=\frac{E_{n}\lambda }{hc}\)
Where, \(\lambda\)= wavelength of light = 4000 pm = \(4000\times 10^{12}\, m\)
c = velocity of light in vacuum =\(3\times 10^{8}\, m/s\)
h = Planck’s constant = \(6.626\times 10^{34}\, Js\)
Substituting the values in the given expression of n:
\(n=\frac{1\times (4000\times 10^{12})}{(6.626\times 10^{34})(3\times 10^{8})}=2.012\times 10^{16}\)Hence, the number of photons with a wavelength of 4000 pm and energy of 1 J are \(2.012\times 10^{16}\)
Q.9. A photon of wavelength \(4\times 10^{7}\, m\) strikes on metal surface, the work function of the metal being 2.13 eV. Calculate
(I)the energy of the photon (eV),
(II)the kinetic energy of the emission, and
(III)the velocity of the photoelectron (1 eV= \(1.6020\times 10^{19}\,J\))
Ans: Energy (E) of a photon = \(h\nu =\frac{hc}{\lambda }\)
Where, h = Planck’s constant =\(6.626\times 10^{34}\,Js\)
c = velocity of light in vacuum =\(3\times 10^{8}\,m/s\) \(\lambda\)= wavelength of photon =\(4\times 10^{7}\,m/s\)
Substituting the values in the given expression of E:
\(E=\frac{(6.626\times 10^{34})(3\times 10^{8})}{4\times 10^{7}}=4.9695\times 10^{19}\, J\)Hence, the energy of the photon is \(4.97\times 10^{19}\, J\)
(II) The kinetic energy of emission \(E_{k}\) is given by
\(=h\nu h\nu _{0}\\ \\ =(EW)eV\\ \\ =(\frac{4.9695\times 10^{19}}{1.6020\times 10^{19}})eV2.13\, eV\\ \\ =(3.10202.13)eV\\ \\ =0.9720\, eV\)Hence, the kinetic energy of emission is 0.97 eV.
(III)The velocity of a photoelectron (v) can be calculated by the expression,
\(\frac{1}{2}mv^{2} =h\nu h\nu _{0}\\ \\ \Rightarrow v =\sqrt{\frac{2(h\nu h\nu _{0})}{m}}\)Where \((h\nu h\nu _{0})\) is the kinetic energy of emission in Joules and ‘m’ is the mass of the photoelectron.
Substituting the values in the given expression of v:
\(v=\sqrt{\frac{2\times (0.9720\times 1.6020\times 10^{19})J}{9.10939\times 10^{31}kg}}\\ \\ =\sqrt{0.3418\times 10^{12}m^{2}s^{2}}\\ \\ \Rightarrow v=5.84\times 10^{5}ms^{1}\)Hence, the velocity of the photoelectron is \(5.84\times 10^{5}ms^{1}\)
Q.10.Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ \(mol^{1}\).
Ans: Energy of sodium (E)= \(\frac{N_{A}hc}{\lambda }\\ \\ =\frac{(6.023\times 10^{23}\, mol^{1})(6.626\times 10^{34})Js(3\times 10^{8})ms^{1}}{242\times 10^{9}m}\\ \\ =4.947\times 10^{5}\, J\, mol^{1}\\ \\ =494.7\times 10^{3}\, J\, mol^{1}\\ \\ =494\, kJ\, mol^{1}\)
Q.11. A 25 watt bulb discharges monochromatic yellow light of wavelength of 0.57µm. Ascertain the rate of discharge of quanta every second.
Ans: Power of bulb, P = 25 Watt = \(25\, Js^{1}\)
Energy of one photon, E = \(h\nu =\frac{hc}{\nu }\)
Substituting the values in the given expression of E:
\(E=\frac{(6.626\times 10^{34})(3\times 10^{8})}{(0.57\times 10^{6})}=34.87\times 10^{20}\, J\)E= \(34.87\times 10^{20}\, J\)
Rate of emission of quanta per second= \(E=\frac{25}{34.87\times 10^{20}}=7.169\times 10^{19}\, s^{1}\)
Q.12. Electrons are discharged with zero speed from a metal surface when it is presented to radiation of wavelength 6800 Å. Figure limit recurrence (\(\nu _{0}\) ) and work work (\(W_{0}\)) of the metal.
Ans: Threshold wavelength of radiation \((\lambda _{0})\)= 6800 Å=\(6800\times 10^{10}\, m\)
Threshold frequency (\(\nu _{0}\) ) of the metal
=\(\frac{c}{\lambda _{0}}=\frac{3\times 10^{8}ms^{1}}{6.8\times 10^{7}m}=4.41\times 10^{14}\, s^{1}\)
Thus, the threshold frequency (\(\nu _{0}\) ) of the metal = \(h\nu _{0}\\ \\ =(6.626\times 10^{34}Js)(4.41\times 10^{14}s^{1})\\ \\ =2.922\times 10^{19}\, J\)
Q.13. What is the wavelength of light transmitted when the electron in a hydrogen atom experiences move from an energy level with n = 4 to an energy level with n = 2?
Ans: The \(n_{i}\) = 4 to \(n_{f}\) = 2 transition will give rise to a spectral line of the Balmer series. The energy involved in the transition is given by the relation,
\(E=2.18\times 10^{18}[\frac{1}{n_{i}^{2}}\frac{1}{n_{f}^{2}}]\)Substituting the values in the given expression of E:
\(E=2.18\times 10^{18}[\frac{1}{4^{2}}\frac{1}{2^{2}}]\\ \\ =2.18\times 10^{18}[\frac{14}{16}]\\ \\ =2.18\times 10^{18}\times (\frac{3}{16})\\ \\ E=(4.0875\times 10^{19}J)\)The negative sign indicates the energy of emission.
Wavelength of light emitted \((\lambda )=\frac{hc}{E}\)
(Since \(E=\frac{hc}{\ lambda }\))
Substituting the values in the given expression of \({\ lambda }\):
\(\lambda =\frac{(6.626\times 10^{34})(3\times 10^{8})}{(4.0875\times 10^{19})}=4.8631\times 10^{7}\, m\\ \\ \lambda =486.31\times 10^{9}\, m\\ \\ =486\, nm\)
Q.14. What amount of energy is required to ionize a H molecule if the electron involves n = 5 circle? Contrast your answer and the ionization enthalpy of H molecule (energy required to evacuate the electron from n =1 circle).
Ans: The expression of energy is given by,
\(E_{n} =\frac{(2.18\times 10^{18})Z^{2}}{n^{2}}\)Where,
Z= atomic number of the atom n
= principal quantum number
For ionization from \(n_{t}=5 \) to \(n_{2} =∞\),
\(\Delta E=E_{\infty }E_{5 }\\ \\ =[(\frac{(2.18\times 10^{18}\, J)(1)^{2}}{(\infty )^{2}})(\frac{(2.18\times 10^{18}\, J)(1)^{2}}{(5 )^{2}})]\\ \\ =0.0872\times 10^{18}\, J\\ \\ \Delta E=8.72\times 10^{20}\, J\)Hence, the energy required for ionization from n = 5 to n = ∞Energy required for n1 = 1 to n = ∞, is \(8.72\times 10^{20}\, m\)
\(\Delta E=E_{\infty }E_{5 }\\ \\ =[(\frac{(2.18\times 10^{18}\, J)(1)^{2}}{(\infty )^{2}})(\frac{(2.18\times 10^{18}\, J)(1)^{2}}{(1 )^{2}})]\\ \\ =2.18\times 10^{18}\, J\\ \\ \Delta E=2.18\times 10^{18}\, J\)
Hence, less energy is required to ionize an electron in the 5th orbital of hydrogen atom as compared to that in the ground state.
Q.15. When the excited electron of an H atom in n = 6 drops to the ground state, the following transitions are possible:
Hence, a total number of (5 + 4 + 3 + 2 + 1) 15 lines will be obtained in the emission spectrum.
The number of spectral lines produced when an electron in the n th level drops down to the ground state is given by
\(\frac{n(n1)}{2}\)Given, n = 6
Number of spectral lines =\(\frac{6(61)}{2}=15\)
Q.16.(I) Energy associated with the fifth orbit of hydrogen atom is calculated as:
\(E_{5}=\frac{(2.18\times 10^{18})}{(5)^{2}}=\frac{(2.18\times 10^{18})}{25}=8.72\times 10^{20}\, J\)
(II) Radius of Bohr’s n th orbit for hydrogen atom is given by, r_{n} = (0.0529 nm) n ^{2}
For, n = 5
\(r_{5}=(0.0529\, nm)(5^{2})\\ \\ r_{5}=1.3225\, nm\)
Q.17. Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.
Ans. For the Balmer series, n_{i} = 2. Thus, the expression of wavenumber(\(\bar{\nu}\)) is given by,
\(\bar{\nu}=[\frac{1}{(2)^{2}}\frac{1}{n_{f}^{2}}](1.097\times 10^{7}\, m^{1})\)Wave number(\(\bar{\nu}\)) is inversely proportional to wavelength of transition. Hence, for the longest wavelength transition, (\(\bar{\nu}\)) has to be the smallest.
For (\(\bar{\nu}\)) to be minimum, n_{f} should be minimum.
For the Balmer series, a transition from n_{i} = 2 to n_{f} = 3 is allowed.
Hence, taking n_{f} = 3, we get:
\(\bar{\nu}=(1.097\times 10^{7})[\frac{1}{(2)^{2}}\frac{1}{{3}^{2}}]\\ \\ \bar{\nu}=(1.097\times 10^{7})[\frac{1}{4}\frac{1}{9}]\\ \\ =(1.097\times 10^{7})[\frac{94}{36}]\\ \\ =(1.097\times 10^{7})[\frac{5}{36}]\\ \\ \bar{\nu}=1.5236\times 10^{6}\, m^{1}\)
Q.18. What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state?
The ground state electron energy is –2.18 × 10^{–11} ergs.
Ans. Energy (E) of the n^{th} Bohr orbit of an atom is given by,
\(E_{5}=\frac{(2.18\times 10^{18})Z^{2}}{(n)^{2}}\)Where, Z = atomic number of the atom
Ground state energy \(=2.18 \times 10^{–11}\, ergs\\ =2.18 \times 10^{–11} \times 10^{–7}\, J \\ =2.18 \times 10^{–18} \, J\)
Energy required to shift the electron from n = 1 to n = 5 is given as:
\(\Delta E=E_{5}E_{1}\\ \\ =[(\frac{(2.18\times 10^{18}\, J)(1)^{2}}{(5 )^{2}})(2.18\times 10^{18})]\\ \\ =(2.18\times 10^{18})[1\frac{1}{25}]\\ \\ =(2.18\times 10^{18})[\frac{24}{25}]\\ \\ =2.0928\times 10^{18}\, J\)Wavelength of emitted light=
\( \frac{hc}{E }\) \(E=\frac{(6.626\times 10^{34})(3\times 10^{8})}{(2.0928\times 10^{18})}\\ \\ =9.498\times 10^{8}\, m\)
Q.19. The electron energy in hydrogen atom is given by E_{n} =\(2.18\times 10^{18}, n^{2}J\). Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?
Ans. \(E_{n}=\frac{(2.18\times 10^{18}) }{(n)^{2} }J\)
Energy required for ionization from n = 2 is given by,
\(\Delta E=E_{\infty }E_{2}\\ \\ =[(\frac{(2.18\times 10^{18}}{(\infty )^{2}})(\frac{(2.18\times 10^{18})}{(2)^{2}})]J\\ \\ =[\frac{(2.18\times 10^{18})}{4}0]J\\ \\ =0.545\times 10^{18}\, J\) \(\Delta E=5.45\times 10^{19}\, J\) \(\lambda =\frac{hc}{\Delta E}\)Here, λ is the longest wavelength causing the transition.
\(E=\frac{(6.626\times 10^{34})(3\times 10^{8})}{(0.57\times 10^{6})}=34.87\times 10^{20}\, J\) \(E=\frac{(6.626\times 10^{34})(3\times 10^{8})}{(5.45\times 10^{19})}=3.647\times 10^{7}\, m\\ \\ =3647\times 10^{10}\)= 3647 Å
Q.20. Calculate the wavelength of an electron moving with a velocity of \(2.05\times 10^{7}\, ms^{1}\)
Ans. According to de Broglie’s equation,
\(\lambda =\frac{h}{mv }\)Where, λ = wavelength of moving particle
m = mass of particle
v = velocity of particle
h = Planck’s constant
Substituting the values in the expression of λ:
\(\lambda =\frac{(6.626\times 10^{34})Js}{(9.10939\times 10^{31}kg)(2.05\times 10^{7}ms^{1})}\\ \\ =3.548\times 10^{11}\, m\)Hence, the wavelength of the electron moving with a velocity of \(2.05\times 10^{7}\, ms^{1}\: is\:\: 3.548\times 10^{11}\, m\)
Q.21. The mass of an electron is \(9.1 \times 10^{31}\)kg. If its K.E. is \(3.0\times 10^{25}\)J, calculate its wavelength.
Ans. From de Broglie’s equation,
\(\lambda =\frac{h}{mv }\)Given, Kinetic energy (K.E) of the electron = \(3.0\times 10^{25}\)J
Since K.E.= \(\frac{1}{2}mv^{2}\) \(∴ Velocity(v)=\sqrt{\frac{2K.E}{m}}\\ \\ =\sqrt{\frac{2(3.0\times 10^{25}J)}{9.10939\times 10^{31}kg}}\\ \\ =\sqrt{6.5866\times 10^{4}}\\ \\ v=811.579\, ms^{1}\)
Substituting the value in the expression of λ:
\(\lambda =\frac{(6.626\times 10^{34})Js}{(9.10939\times 10^{31}kg)(811.579ms^{1})}\\ \\ =8.9625\times 10^{7}\, m\)
Q.22. (I)Write the electronic configurations of the following ions:
(a)H^{–}
(b)Na^{+}
(c)O^{2–}
(d)F^{ –}
(II) What are the atomic numbers of elements whose outermost electrons are represented by
(a)3s^{1}
(b)2p^{3} and
(c)3p^{5}?
(III) Which atoms are indicated by the following configurations?
(a)[He] 2s^{1}
(b)[Ne] 3s^{2} 3p^{3}
(c)[Ar] 4s^{2} 3d^{1} .
Ans:
(I)(a)H^{–} ion
The electronic configuration of H atom is 1s^{1} .
A negative charge on the species indicates the gain of an electron by it.
\(∴\)Electronic configuration of H^{–} = 1s^{2}
(b)Na^{+ }ion
The electronic configuration of Na atom is 1s^{2} 2s^{2 }2p^{6} 3s^{1} .
A positive charge on the species indicates the loss of an electron by it.
\(∴\)Electronic configuration of Na^{+} = 1s^{2} 2s^{2} 2p^{6} 3s^{0} or 1s^{2} 2s^{2} 2p^{6}(c)O^{2–} ion
The electronic configuration of 0 atom is 1s ^{2} 2s ^{2} 2p^{ 4} .
A dinegative charge on the species indicates that two electrons are gained by it. \(∴\)Electronic configuration of O^{2–} ion = 1s ^{2} 2s ^{2 }p ^{6}
(d)F^{ –} ion
The electronic configuration of F atom is 1s^{ 2} 2s ^{2} 2p ^{5} .
A negative charge on the species indicates the gain of an electron by it.
\(∴\)Electron configuration of F^{–} ion = 1s^{ 2} 2s ^{2} 2p ^{6}(II) (a)3s^{1}
Completing the electron configuration of the element as 1s ^{2} 2s^{ 2} 2p ^{6} 3s ^{1} . Number of electrons present in the atom of the element = 2 + 2 + 6 + 1 = 11 \(∴\)Atomic number of the element = 11
(b)2p ^{3}
Completing the electron configuration of the element as 1s ^{2} 2s ^{2} 2p ^{3} .
Number of electrons present in the atom of the element = 2 + 2 + 3 = 7
\(∴\)Atomic number of the element = 7(c)3p ^{5}
Completing the electron configuration of the element as 1s ^{2} 2s ^{2} 2p ^{5} .
Number of electrons present in the atom of the element = 2 + 2 + 5 = 9
\(∴\)Atomic number of the element = 9(III)(a)[He] 2s ^{1}
The electronic configuration of the element is [He] 2s ^{1} = 1s ^{2} 2s ^{1} .
\(∴\)Atomic number of the element = 3Hence, the element with the electronic configuration [He] 2s ^{1} is lithium (Li).
(b)[Ne] 3s ^{2} 3p^{3}
The electronic configuration of the element is [Ne] 3s^{ 2} 3p ^{3}= 1s ^{2 }2s ^{2} 2p ^{6} 3s ^{2 }3p^{ 3} . \(∴\)Atomic number of the element = 15
Hence, the element with the electronic configuration [Ne] 3s ^{2} 3p ^{3} is phosphorus (P).
(c)[Ar] 4s ^{2} 3d^{1}
The electronic configuration of the element is [Ar] 4s ^{2} 3d^{ 1}= 1s ^{2} 2s ^{2 }2p ^{6} 3s ^{2} 3p ^{6} 4s ^{2 }3d^{ 1} .
\(∴\)Atomic number of the element = 21Hence, the element with the electronic configuration [Ar] 4s ^{2} 3d ^{1} is scandium (Sc).
Q.23. What is the most reduced estimation of n that permits g orbitals to exist?
Ans. For gorbitals, l = 4.
With respect to any value “n” of principal quantum number, the Azimuthal quantum number (l) can have a value from zero to (n – 1).
\(∴\)For l = 4, least value of n = 5.
Q.24. An electron is in one of the 3d orbitals. Give the possible values of n_{l}, l and m_{l} for this electron.
Ans: For the 3d orbital:
Principal quantum number (n) = 3
Azimuthal quantum number (l) = 2
Magnetic quantum number (m_{l}) = – 2, – 1, 0, 1, 2
Q.25.An atom of an element contains 29 electrons and 35 neutrons. Deduce
(i)The number of protons and
(ii)The electronic configuration of the element.
Ans:
(i)For an atom to be neutral, the number of protons is equal to the number of electrons.
\(∴\)Number of protons in the atom of the given element = 29(ii)The electronic configuration of the atom is 1s ^{2} 2s ^{2} 2p ^{6} 3s^{ 2} 3p^{ 6} 4s ^{2} 3d ^{10} .
Q.26.Give the number of electrons in the species , H_{2}^{+} and H_{2} and O_{2}+
Ans: Number of electrons present in hydrogen molecule (H_{2}^{+}) = 1 + 1 = 2
\(∴\)Number of electrons in = 2 – 1 = 1H_{2}: Number of electrons in H2 = 1 + 1 = 2
O_{2}+:Number of electrons present in oxygen molecule (O_{2}) = 8 + 8 = 16
\(∴\)Number of electrons in = 16 – 1 = 15
Q.27. (I)An atomic orbital has n = 3. What are the possible values of l and m_{l} ?
(II)List the quantum numbers (m_{l }and l) of electrons for 3d orbital.
(III) Which of the following orbitals are possible? 1p, 2s, 2p and 3f
Ans.
(I)n = 3 (Given)
For a given value of n, l can have values from 0 to (n – 1).
\(∴\)For n = 3l = 0, 1, 2
For a given value of l, m_{l} can have (2l + 1) values.
For l = 0, m = 0 l = 1, m = – 1, 0, 1 l = 2, m = – 2, – 1, 0, 1, 2
For n = 3
l = 0, 1, 2
m_{0 }= 0
m_{1} = – 1, 0, 1
m_{2} = – 2, – 1, 0, 1, 2
(II)For 3d orbital, l = 2.
For a given value of l, ml can have (2l + 1) values i.e., 5 values.
For l = 2 m_{2 }= – 2, – 1, 0, 1, 2
(III)Among the given orbitals only 2s and 2p are possible. 1p and 3f cannot exist. For porbital, l = 1.
For a given value of n, l can have values from zero to (n – 1).
For l is equal to 1, the minimum value of n is 2.
Similarly,
For forbital, l = 4.
For l = 4, the minimum value of n is 5.
Hence, 1p and 3f do not exist.
Q.28.Using s, p, d notations, describe the orbital with the following quantum numbers.
(a)n = 1, l = 0;
(b)n = 3; l =1
(c) n = 4; l = 2;
(d) n = 4; l =3.
Ans:
(a)n = 1, l = 0 (Given) The orbital is 1s.
(b)For n = 3 and l = 1 The orbital is 3p.
(c)For n = 4 and l = 2 The orbital is 4d.
(d)For n = 4 and l = 3 The orbital is 4f.
Q.29. Explain, giving reasons, which of the following sets of quantum numbers are not possible.
a) n = 0, l = 0, m_{l}= 0, m_{s }=+\(\frac{1}{2}\)
b) n = 1, l = 0, m_{l}= 0, m_{s }=\(\frac{1}{2}\)
c) n = 1, l = 1, m_{l}= 0, m_{s }=+\(\frac{1}{2}\)
d) n = 2, l = 0, m_{l}= 1, m_{s }=\(\frac{1}{2}\)
e) n = 3, l = 3, m_{l}= 3, m_{s }=+\(\frac{1}{2}\)
f) n = 3, l = 0, m_{l}= 1, m_{s }=+\(\frac{1}{2}\)
Ans.(a)The given set of quantum numbers is not possible because the value of the principal quantum number (n) cannot be zero.
(b)The given set of quantum numbers is possible.
(c)The given set of quantum numbers is not possible. For a given value of n, ‘l’ can have values from zero to (n – 1). For n = 1, l = 0 and not 1.
(d)The given set of quantum numbers is possible. (e) The given set of quantum numbers is not possible. For n = 3, l = 0 to (3 – 1) l = 0 to 2 i.e., 0, 1, 2
(f)The given set of quantum numbers is possible.
Q.30. How many electrons in an atom may have the following quantum numbers?
a) n = 4, m_{s }=\(\frac{1}{2}\)
b)n = 3, l = 0
Ans.(a)Total number of electrons in an atom for a value of n = 2n ^{2}
\(∴\)For n = 4,Total number of electrons = 2 (4)^{2} = 32
The given element has a fully filled orbital as
1s ^{2} 2s ^{2} 2p ^{6} 3s ^{2} 3p ^{6} 4s^{ 2} 3d ^{10} .
Hence, all the electrons are paired.
\(∴\)Number of electrons (having n = 4 and m_{s }=\(\frac{1}{2}\)) = 16(b)n = 3, l = 0 indicates that the electrons are present in the 3s orbital. Therefore, the number of electrons having n = 3 and l = 0 is 2.
Q.31. Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.
Ans.Since a hydrogen atom has only one electron, according to Bohr’s postulate, the angular momentum of that electron is given by:
\(mvr=n\frac{h}{2\pi }\: \: \: ….(1)\)Where, n = 1, 2, 3, …
According to de Broglie’s equation:
\(\lambda =\frac{h}{mv }\\ \\ or\: mv=\frac{h}{\lambda }\: \: \: \: …….(2)\) \(\frac{hr}{\lambda }=n\frac{h}{2\pi }\\ \\ or\: 2\pi r=n\lambda \: \: \: \: …..(3)\)Since ‘2πr’ represents the circumference of the Bohr orbit (r), it is proved by equation (3) that the circumference of the Bohr orbit of the hydrogen atom is an integral multiple of de Broglie’s wavelength associated with the electron revolving around the orbit.
Q.32.What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He^{+ }spectrum?
Ans. For He^{+} ion, the wave number associated with the Balmer transition, n = 4 to n = 2 is given by:
\(\bar{\nu}=\frac{1}{\lambda }=RZ^{2}(\frac{1}{n_{1}^{2}}\frac{1}{n_{2}^{2}})\)Where, n_{1} = 2 n_{2} = 4
Z = atomic number of helium
\(\bar{\nu}=\frac{1}{\lambda }=R(2)^{2}(\frac{1}{4}\frac{1}{16})\\ \\ =4R(\frac{41}{16})\\ \\ \bar{\nu}=\frac{1}{\lambda }=\frac{3R}{4}\\ \\ \Rightarrow \lambda =\frac{4}{3R}\)According to the question, the desired transition for hydrogen will have the same wavelength as that of He^{+}.
\(R(1)^{2}[\frac{1}{n_{1}^{2}}\frac{1}{n_{2}^{2}}]=\frac{3R}{4}\\ \\ \Rightarrow [\frac{1}{n_{1}^{2}}\frac{1}{n_{2}^{2}}]=\frac{3}{4}\: \: \: \: \: …….(1)\)By hit and trail method, the equality given by equation (1) is true only when n_{1} = 1and n_{2} = 2.
The transition for n_{2} = 2 to n = 1 in hydrogen spectrum would have the same wavelength as Balmer transition n = 4 to n = 2 of He^{+} spectrum.
Q.33.Calculate the energy required for the process
\(He^{+}_{(g)}\rightarrow He^{2+}_{(g)}+e^{}\)
The ionization energy for the H atom in the ground state is \(2.18\times 10^{18}\, J\,atom^{1}\)
Ans. Energy associated with hydrogenlike species is given by,
\(E_{n}=2.18\times 10^{18}(\frac{Z^{2}}{n^{2}})J\)For ground state of hydrogen atom,
\(\Delta E=E_{\infty }E_{1}\\ \\ =0[(2.18\times 10^{18}{\frac{(1)^{2}}{(1)^{2}}}]J\\ \\ \Delta E=2.18\times 10^{18}J\\\)For the given process,
\(He^{+}_{(g)}\rightarrow He^{2+}_{(g)}+e^{}\)An electron is removed from n = 1 to n = ∞.
\(\Delta E=E_{\infty }E_{1}\\ \\ =0[(2.18\times 10^{18}{\frac{(2)^{2}}{(1)^{2}}}]J\\ \\ \Delta E=8.72\times 10^{18}J\\\). \(∴\) The energy required for the process \(8.72\times 10^{18}J\\\).
Q.34. If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long.
Ans. 1 m = 100 cm
1 cm = 10^{–2} m
Length of the scale = 20 cm
=\(20\times 10^{2}\, m\)
Diameter of a carbon atom = 0.15 nm
=\(0.15\times 10^{9}\, m\)
One carbon atom occupies \(0.15\times 10^{9}\, m\).
\(∴\) Number of carbon atoms that can be placed in a straight line \(\frac{20\times 10^{2}m}{0.15\times 10^{9}m}\\ \\ =133.33\times 10^{7}\\ \\ =1.33\times 10^{9}\)
Q.35. \(2\times 10^{8}\, m\) atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm.
Ans. Length of the given arrangement = 2.4 cm
Number of carbon atoms present = 2 × 10^{8}
Diameter of carbon atom
\(\frac{2.4\times 10^{2}m}{2\times 10^{8}m}\\ \\ =1.2\times 10^{10}m\) \(∴\)Radius of carbon atom=\(\frac{Diameter}{2}\\ \\ =\frac{1.2\times 10^{10}m}{2}\\ \\ =6.0\times 10^{11}m\)
Q.36.The diameter of zinc atom is 2.6Å.Calculate (a) radius of zinc atom in pm and (b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise.
Ans. (a)Radius of carbon atom= \(\frac{Diameter}{2}\\ \\ =\frac{2.6}{2}\\ \\ =1.3\times 10^{10}m\\ \\ =130\times 10^{12}m=130pm\)
(b) )Length of the arrangement = 1.6 cm
=\( 1.6 \times 10^{2}m\)
Diameter of zinc atom =\( 1.6 \times 10^{10}m\) \(∴\) Number of zinc atoms present in the arrangement
\(=\frac{1.6\times 10^{2}m}{2.6\times 10^{10}m}\\ \\ =0.6153\times 10^{8}m\\ \\ =6.153\times 10^{7}\)
Q.37. The diameter of zinc atom is 2.6Å.Calculate (a) radius of zinc atom in pm and (b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise.
Ans.
(a)Radius of carbon atom= \(\frac{Diameter}{2}\\ \\ =\frac{2.6}{2}\\ \\ =1.3\times 10^{10}m\\ \\ =130\times 10^{12}m=130pm\)
(b) )Length of the arrangement = 1.6 cm
=\( 1.6 \times 10^{2}m\)
Diameter of zinc atom =\( 1.6 \times 10^{10}m\)
Therefore, Number of zinc atoms present in the arrangement
\(=\frac{1.6\times 10^{2}m}{2.6\times 10^{10}m}\\ \\ =0.6153\times 10^{8}m\\ \\ =6.153\times 10^{7}\)
Q.38. A certain particle carries \( 2.5 \times 10^{16}C\) of static electric charge. Calculate the number of electrons present in it.
Ans.
Charge on one electron = \( 1.6022 \times 10^{19}C\) \(\Rightarrow 1.6022 \times 10^{19}C\) charge is carried by 1 electron.
Therefore, Number of electrons carrying a charge of \( 2.5 \times 10^{16}C\) \(\frac{1}{1.6022\times 10^{19}C}(2.5\times 10^{16}C)\\ \\ =1.560\times 10^{3}C\\ \\ =1560\: C\)
Q.39. In Milikan’s experiment, static electric charge on the oil drops has been obtained by shining Xrays. If the static electric charge on the oil drop is\( 1.282 \times 10^{18}C\) , calculate the number of electrons present on it.
Ans.
Charge on the oil drop =\( 1.282 \times 10^{18}C\)
Charge on one electron =\( 1.6022 \times 10^{19}C\)
Therefore, Number of electrons present on the oil drop
\(\frac{1.282\times 10^{18}C}{1.6022\times 10^{19}C}\\ \\ =0.8001\times 10^{1}\\ \\ =8.0\)
Q.40. In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been used to be bombarded by the αparticles. If the thin foil of light atoms like Aluminium etc. is used, what difference would be observed from the above results?
Ans.
A thin foil of lighter atoms will not give the same results as given with the foil of heavier atoms.
Lighter atoms would be able to carry very little positive charge. Hence, they will not cause enough deflection of αparticles (positively charged).
Q.41. Symbols \(_{35}^{79}\textrm{Br}\: and\: _{}^{79}\textrm{Br}\) can be written, whereas symbols \(_{79}^{35}\textrm{Br}\: and\: _{}^{35}\textrm{Br}\) are not acceptable. Answer briefly.
Ans.
The general convention of representing an element along with its atomic mass (A) and atomic number (Z) is \(_{Z}^{A}\textrm{X}\).
Hence, \(_{35}^{79}\textrm{Br}\) is acceptable but \(_{79}^{35}\textrm{Br}\) is not acceptable.
\( _{}^{79}\textrm{Br}\) can be written but \( _{}^{35}\textrm{Br}\) cannot be written because the atomic number of an element is constant, but the atomic mass of an element depends upon the relative abundance of its isotopes. Hence, it is necessary to mention the atomic mass of an element.
Q.42. An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.
Ans.
Let the number of protons in the element be x.
Therefore, Number of neutrons in the element = x + 31.7% of x
= x + 0.317 x
= 1.317 x
According to the question,
Mass number of the element = 81
Therefore, (Number of protons + number of neutrons) = 81
\(\Rightarrow x+1.317x=81\\ \\ 2.317x=81\\ \\ x=\frac{81}{2.317}\\ \\ =34.95\\ \\\) \(x\simeq 35\)Hence, the number of protons in the element i.e., x is 35.
Since the atomic number of an atom is defined as the number of protons present in its nucleus, the atomic number of the given element is 35.
Therefore, the atomic symbol of the element is \(_{35}^{81}\textrm{Br}\)
Q.43. An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than the electrons, find the symbol of the ion.
Ans.
Let the number of electrons in the ion carrying a negative charge be x.
Then,
Number of neutrons present = x + 11.1% of x
= x + 0.111 x
= 1.111 x
Number of electrons in the neutral atom = (x – 1)
(When an ion carries a negative charge, it carries an extra electron)
Therefore, Number of protons in the neutral atom = x – 1
Given,
Mass number of the ion = 37
Therefore, (x – 1) + 1.111x = 37
2.111x = 38 x
= 18
Therefore, The symbol of the ion is \(_{17}^{37}\textrm{Cl}^{}\)
Q.44.An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol to this ion.
Ans.
Let the number of electrons present in ion \(A^{3+}\) be x.
Therefore, Number of neutrons in it = x + 30.4% of x = 1.304 x
Since the ion is tripositive,
\(\Rightarrow\) Number of electrons in neutral atom = x + 3Therefore, Number of protons in neutral atom = x + 3
Given, Mass number of the ion = 56
Therefore, (x+3)(1.304x)=56
2.304x=53
\(x=\frac{53}{2.304}\)X=23
Therefore, Number of protons = x + 3 = 23 + 3 = 26
Therefore, The symbol of the ion \(_{26}^{56}\textrm{Fe}^{+}\)
Q.45. Arrange the following type of radiations in increasing order of frequency: (a) radiation from microwave oven (b) amber light from traffic signal (c) radiation from FM radio (d) cosmic rays from outer space and (e) Xrays.
Ans.
The increasing order of frequency is as follows:
Radiation from FM radio < amber light < radiation from microwave oven < X rays < cosmic rays
The increasing order of wavelength is as follows:
Cosmic rays < Xrays < radiation from microwave ovens < amber light < radiation of FM radio
Q.46. Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is \(5.6 \times 10^{24}\), calculate the power of this laser.
Ans.
Power of laser = Energy with which it emits photons
Power=\(=E=\frac{Nhc}{\lambda }\)
Where, N = number of photons emitted
h = Planck’s constant
c = velocity of radiation
λ = wavelength of radiation
Substituting the values in the given expression of Energy (E):
\(E=\frac{(5.6\times 10^{24})(6.626\times 10^{34}Js)(3\times 10^{8}ms^{1})}{(337.1\times 10^{9}m) }\\ \\ =0.3302\times 10^{7}J\\ \\ =3.33\times 10^{6}J\)Hence, the power of the laser is \(3.33\times 10^{6}J\)
Q.47. Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate (a) the frequency of emission,
(b) distance traveled by this radiation in 30 s
(c) energy of quantum and
(d) number of quanta present if it produces 2 J of energy.
Ans.
Wavelength of radiation emitted = 616 nm =\(616\times 10^{9}m\)(Given)
(a)Frequency of emission (\(\nu\))
\(\nu =\frac{c}{\lambda }\)Where, c = velocity of radiation
λ = wavelength of radiation
Substituting the values in the given expression of (\(\nu\)):
\(\nu =\frac{3\times 10^{8}m/s}{616\times 10^{9} m}\\ \\ =4.87\times 10^{8}\times 10^{9}\times 10^{3}s^{1}\, v\\ \\ =4.87\times 10^{14}s^{1}\)Frequency of emission (\(\nu\))= \(4.87\times 10^{14}s^{1}\)
(b) Velocity of radiation,c =\(3\times 10^{8}ms^{1}\)
Distance travelled by this radiation in 30 s
\(=(3\times 10^{8}ms^{1})(30s)\\ \\ =9\times 10^{9}m\)(c) Energy of quantum (E) = hν
\(=(6.626\times 10^{34}Js)(4.87\times 10^{14}s^{1})\)Energy of quantum (E) =\(32.27\times 10^{20}J\)
Therefore, \(32.27\times 10^{20}J\) of energy is present in 1 quantum.
Number of quanta in 2 J of energy
\(\frac{2J}{32.27\times 10^{20}J}\\ \\ =6.19\times 10^{18}\\ \\ =6.2\times 10^{18}\)
Q.48. In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of \(3.15\times 10^{18}J\) from the radiations of 600 nm, calculate the number of photons received by the detector.
Ans.
From the expression of energy of one photon (E),
\(E=\frac{hc}{\lambda }\)Where,
λ = wavelength of radiation
h = Planck’s constant
c = velocity of radiation
Substituting the values in the given expression of E:
\(E=\frac{(6.626\times 10^{34}Js)(3\times 10^{8}ms^{1})}{(600\times 10^{9})}=3.313\times 10^{19}\, J\)Energy of one photon =\(3.313\times 10^{19}\, J\)
Number of photons received with \(3.15\times 10^{18}\, J\)energy
\(=\frac{3.15\times 10^{18}J}{3.313\times 10^{19}J }\\ \\ =9.5\\ \\ \approx 10\)
Q.49. Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is \(2.5\times 10^{15}\, J\), calculate the energy of the source.
Ans.
Frequency of radiation (\(\nu\)),
\(\nu =\frac{1}{2.0\times 10^{9}s}\\ \\ \nu =5.0\times 10^{8}s^{1}\)Energy (E) of source = Nhν
Where,
N = number of photons emitted
h = Planck’s constant
ν = frequency of radiation
Substituting the values in the given expression of (E):
\(E=(2.5\times 10^{15})(6.626\times 10^{34}Js)(5.0\times 10^{8}s^{1})\\ \\ E=8.282\times 10^{10}J\)Hence, the energy of the source (E) is \(8.282\times 10^{10}J\).
Q.51. The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.
Ans.
It is given that the work function (W_{0}) for caesium atom is 1.9 eV.
(a)From the \(W_{0}=\frac{hc}{\lambda _{0}}\) expression, we get:
\(\lambda _{0}=\frac{hc}{W _{0}}\)Where,
λ_{0} = threshold wavelength
h = Planck’s constant
c = velocity of radiation
Substituting the values in the given expression of (λ_{0}):
\(\lambda _{0}=\frac{(6.626\times 10^{34}Js)(3\times 10^{8}ms^{1})}{(1.9\times1.602\times 10^{19}J)}=6.53\times 10^{7}\, m\)Hence, the threshold wavelength (λ_{0}) is 653 nm.
(b) From the expression, \(W_{0}=h\nu _{0}\) , we get:
\(\nu _{0}=\frac{W_{0}}{h}\)Where,
ν_{0} = threshold frequency
h = Planck’s constant
Substituting the values in the given expression of ν_{0}:
\(\nu _{0}=\frac{1.9\times1.602\times 10^{19}J}{6.626\times 10^{34}Js}\)(1 eV = \(1.602\times 10^{19}\, J\))
ν_{0} = \(4.593\times 10^{14}\, s^{1}\)
Hence, the threshold frequency of radiation (ν_{0}) is \(4.593\times 10^{14}\, s^{1}\)
(c) According to the question:
Wavelength used in irradiation (λ) = 500 nm
Kinetic energy = h (ν – ν_{0})
\(=hc(\frac{1}{\lambda }\frac{1}{\lambda _{0}})\\ \\ =(6.626\times 10^{34}Js)(3.0\times 10^{8}ms^{1})(\frac{\lambda _{0}\lambda }{\lambda \lambda _{0}})\\ \\ =(1.9878\times 10^{26}Jm)[\frac{(653500)10^{9}m}{(653)(500)10^{18}m^{2}}]\\ \\ =\frac{(1.9878\times 10^{26})(153\times 10^{9})}{(653)(500)}J\\ \\ =9.3149\times 10^{20}J\)Kinetic energy of the ejected photoelectron =\(9.3149\times 10^{20}J\)
Since K.E=\(\frac{1}{2}mv^{2}=9.3149\times 10^{20}J\) \(\nu =\sqrt{\frac{2(9.3149\times 10^{20}J)}{9.10939\times 10^{31}}}\\ \\ =\sqrt{2.0451\times 10^{11}m^{2}s^{2}}\\ \\ =4.52\times 10^{5}ms^{1}\)
Q.52. Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and, (b) Planck’s constant.
Ans.
(a) Assuming the threshold wavelength to be \(\lambda _{0}\, nm(=\lambda _{0}\times 10^{9}m)\) , the kinetic energy of the radiation is given as:
\(h(\nu \nu _{0})=\frac{1}{2}m\nu ^{2}\)Three different equalities can be formed by the given value as:
\(h(\frac{1}{\lambda } \frac{1}{\lambda _{0}})=\frac{1}{2}m\nu ^{2}\) \(hc(\frac{1}{500\times 10^{9} } \frac{1}{\lambda _{0}\times 10^{9}m})=\frac{1}{2}m(2.55\times 10^{+5}\times 10^{2}ms^{1})\) \(\frac{hc}{10^{9}m}(\frac{1}{500 } \frac{1}{\lambda _{0} })=\frac{1}{2}m(2.55\times 10^{+3}\: ms^{1})^{2}\) (1)Similarly,
\(\frac{hc}{10^{9}m}(\frac{1}{450 } \frac{1}{\lambda _{0} })=\frac{1}{2}m(3.45\times 10^{+3}\: ms^{1})^{2}\) (2) \(\frac{hc}{10^{9}m}(\frac{1}{400 } \frac{1}{\lambda _{0} })=\frac{1}{2}m(5.35\times 10^{+3}\: ms^{1})^{2}\) (3)Dividing equation (3) by equation (1):
\(\frac{[\frac{\lambda _{0}400}{400\lambda _{0}}]}{[\frac{\lambda _{0}500}{500\lambda _{0}}]}=\frac{(5.35\times 10^{+3}ms^{1})^{2}}{(2.55\times 10^{+3}ms^{1})^{2}}\) \(\frac{5\lambda _{0}2000}{4\lambda _{0}2000}=\frac{(5.35)^{2}}{(2.55)^{2}}=\frac{28.6225}{6.5025}\\ \\ \frac{5\lambda _{0}2000}{4\lambda _{0}2000}=4.40177\\ \\ 17.6070\lambda _{0}5\lambda _{0}=8803.5372000\\ \\ \lambda _{0}=\frac{6805.537}{12.607}\\ \\ \lambda _{0}=539.8nm\\ \\ \lambda _{0}=540nm\)So, threshold wavelength (\(\lambda _{0}\)) = 540 nm
Note: part (b) of the question is not done due to the incorrect values of velocity given in the question.
\(\frac{5\lambda _{0}2000}{4\lambda _{0}2000}=\frac{(5.35)^{2}}{(2.55)^{2}}=\frac{28.6225}{6.5025}\\ \\ \frac{5\lambda _{0}2000}{4\lambda _{0}2000}=4.40177\\ \\ 17.6070\lambda _{0}5\lambda _{0}=8803.5372000\\ \\ \lambda _{0}=\frac{6805.537}{12.607}\\ \\ \lambda _{0}=539.8nm\\ \\ \lambda _{0}=540\)
Q.53. The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.
Ans.
From the principle of conservation of energy, the energy of an incident photon (E) is equal to the sum of the work function (W0) of radiation and its kinetic energy (K.E) i.e.,
E = W_{0} + K.E
⇒ W_{0} = E – K.E
Energy of incident photon (E)= \(\frac{hc}{\lambda }\)
Where,
c = velocity of radiation
h = Planck’s constant
λ = wavelength of radiation
Substituting the values in the given expression of E:
\(E=\frac{(6.626\times 10^{34}Js)(3\times 10^{8}ms^{1})}{(256.7\times 10^{9}m)}=7.744\times 10^{19}\, J\\ \\ =\frac{7.744\times 10^{19}}{1.602\times 10^{19}}eV\\ \\ E=4.83\, eV\)The potential applied to silver metal changes to kinetic energy (K.E) of the photoelectron.
Hence,
K.E = 0.35 V
K.E = 0.35 eV
Therefore, Work function, W0 = E – K.E
= 4.83 eV – 0.35 eV
= 4.48 eV
\(\frac{5\lambda _{0}2000}{4\lambda _{0}2000}=\frac{(5.35)^{2}}{(2.55)^{2}}=\frac{28.6225}{6.5025}\\ \\ \frac{5\lambda _{0}2000}{4\lambda _{0}2000}=4.40177\\ \\ 17.6070\lambda _{0}5\lambda _{0}=8803.5372000\\ \\ \lambda _{0}=\frac{6805.537}{12.607}\\ \\ \lambda _{0}=539.8nm\\ \\ \lambda _{0}\simeq 540nm\)
Q.54. If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of \(1.5 \times 10^{7}\, ms^{1}\), calculate the energy with which it is bound to the nucleus.
Ans.
Energy of incident photon (E) is given by,
\(E=\frac{hc}{\lambda }\\ \\ E=\frac{(6.626\times 10^{34})(3\times 10^{8})}{(150\times 10^{12})}=1.3252\times 10^{15}\, J \\ \\ \simeq 13.252\times 10^{16}J\)Energy of the electron ejected (K.E)
\(\frac{1}{2}m_{e}\nu ^{2}\\ \\ =\frac{1}{2}(9.10939\times 10^{31}kg)(1.5\times 10^{7}ms^{1})^{2}\\ \\ =10.2480\times 10^{17}J\\ \\ =1.025\times 10^{16}J\)Hence, the energy with which the electron is bound to the nucleus can be obtained as:
= E – K.E
\(=13.252\times 10^{16}J1.025\times 10^{16}J\\ \\ =12.227\times 10^{16}J\) \(=\frac{12.227\times 10^{16}}{1.602\times 10^{19}}eV\\ \\ =7.6\times 10^{3}eV\) \(\frac{5\lambda _{0}2000}{4\lambda _{0}2000}=\frac{(5.35)^{2}}{(2.55)^{2}}=\frac{28.6225}{6.5025}\\ \\ \frac{5\lambda _{0}2000}{4\lambda _{0}2000}=4.40177\\ \\ 17.6070\lambda _{0}5\lambda _{0}=8803.5372000\\ \\ \lambda _{0}=\frac{6805.537}{12.607}\\ \\ \lambda _{0}=539.8nm\\ \\ \lambda _{0}\simeq 540nm\)
Q.55. Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as v = \(3.29 \times 10^{15}\) (Hz) [1/3^{2} – 1/n ^{2} ] Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.
Ans.
Wavelength of transition = 1285 nm
=\(1285 \times 10^{9} m\)(Given)
\(\nu =3.29\times 10^{15}(\frac{1}{3^{2}}\frac{1}{n^{2}})\)Since \(\nu =\frac{c}{\lambda }\\ \\ =\frac{3\times 10^{8}ms^{1}}{1285\times 10^{9}m}\)
Now, ν = \(2.33 \times 10^{14} s ^{1}\)
Substituting the value of ν in the given expression,
\(3.29\times 10^{15}(\frac{1}{9}\frac{1}{n^{2}})=2.33\times 10^{14}\\ \\ \frac{1}{9}\frac{1}{n^{2}}=\frac{2.33\times 10^{14}}{3.29\times 10^{15}}\\ \\ \frac{1}{9}0.7082\times 10^{1}=\frac{1}{n^{2}} \\ \\ \Rightarrow \frac{1}{n^{2}}=1.1\times 10^{1}0.7082\times 10^{1}\\ \\ \frac{1}{n^{2}}=4.029\times 10^{2}\\ \\ n=\sqrt{\frac{1}{4.029\times 10^{2}}}\)n=4.98 n=5
Hence, for the transition to be observed at 1285 nm, n = 5. The spectrum lies in the infrared region.
Q.56. Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.
Ans.
The radius of the n th orbit of hydrogenlike particles is given by,
\(r=\frac{0.529n^{2}}{Z}\)Å \(r=\frac{5.29n^{2}}{Z}pm\)For radius (r_{1}) = 1.3225 nm
\(=1.32225\times 10^{9}m\\ \\ =1322.25\times 10^{12}m\\ \\ =1322.25\, pm\) \(n_{1}^{2}=\frac{r_{1}Z}{52.9}\\ \\ n_{1}^{2}=\frac{1322.25Z}{52.9}\)Similarly,
\(n_{2}^{2}=\frac{211.6Z}{52.9}\\ \\ \frac{n_{1}^{2}}{n_{2}^{2}}=\frac{1322.5}{211.6}\\ \\ \frac{n_{1}^{2}}{n_{2}^{2}}=6.25\\ \\ \frac{n_{1}}{n_{2}}=2.5\\ \\ \frac{n_{1}}{n_{2}}=\frac{25}{10}=\frac{5}{2}\)⇒ n1 = 5 and n2 = 2 Thus, the transition is from the 5th orbit to the 2nd orbit. It belongs to the Balmer series. wave number (\(\bar{\nu}\))for the transition is given by,
\(1.097\times 10^{7}(\frac{1}{2^{2}}\frac{1}{5^{2}})m^{1}\\ \\ =1.097\times 10^{7}m^{1}(\frac{21}{100})\\ \\ =2.303\times 10^{6}m^{1}\)Wavelength (λ) associated with the emission transition is given by,
\(\lambda =\frac{1}{\nu }\\ \\ =\frac{1}{2.303\times 10^{6}m^{1}}\\ \\ =0.434\times 10^{6}m\, \lambda \\ \\ =434nm\)
Q.57. Dual behavior of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is \(1.6 \times 10^{6}ms^{1}\), calculate de Broglie wavelength associated with this electron.
Ans.
From de Broglie’s equation,
\(\lambda =\frac{h}{m\nu }\\ \\ =\frac{(6.626\times 10^{34})}{9.103939\times 10^{31}kg(1.6\times 10^{6}ms^{1})}\\ \\ =4.55\times 10^{10}m\lambda =455pm\)de Broglie’s wavelength associated with the electron is 455 pm.
Q.58. Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity associated with the neutron.
Ans.
From de Broglie’s equation,
\(\lambda =\frac{h}{m\nu }\\\nu=\frac{h}{m\lambda}\)Where,
v = velocity of particle (neutron)
h = Planck’s constant
m = mass of particle (neutron)
λ = wavelength
Substituting the values in the expression of velocity (v),
\(E=\frac{(6.626\times 10^{34})}{(1.67493\times 10^{27})(800\times 10^{12}m)} =4.94\times 10^{2}\, J \\ =494\, ms^{1}\)Velocity associated with the neutron = 494 ms^{–1}
Q.59. If the velocity of the electron in Bohr’s first orbit is\( 2.19 \times 10^{6} ms^{–1}\) , calculate the de Broglie wavelength associated with it.
Ans.
According to de Broglie’s equation,
\(\lambda =\frac{h}{m\nu }\)Where, λ = wavelength associated with the electron
h = Planck’s constant
m = mass of electron
v = velocity of electron Substituting the values in the expression of λ:
\(\lambda =\frac{h}{m\nu }\\ \\ =\frac{(6.626\times 10^{34})}{9.103939\times 10^{31}kg(2.19\times 10^{6}ms^{1})}\\ \\ =3.32\times 10^{10}m\lambda =332pm\)
Q.60. The velocity associated with a proton moving in a potential difference of 1000 V is\( 4.37 \times 10^{5} ms^{–1}\) . If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity.
Ans.
According to de Broglie’s expression,
\(\lambda =\frac{h}{m\nu }\\ \\ =\frac{(6.626\times 10^{34})}{0.1kg(4.37\times 10^{5}ms^{1})}\\ \\ =1.516\times 10^{38}m\)
Q.61. If the position of the electron is measured within an accuracy of + 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is h/4π_{m} × 0.05 nm, is there any problem in defining this value.
Ans.
From Heisenberg’s uncertainty principle,
Where,
∆x = uncertainty in position of the electron
∆p = uncertainty in momentum of the electron
Substituting the values in the expression of ∆p:
= 2.637 × 10–23 Jsm–1
∆p = 2.637 × 10–23 kgms–1 (1 J = 1 kgms2s –1 )
Uncertainty in the momentum of the electron = 2.637 × 10–23 kgms–1 . = 1.055 × 10–24 kgms–1
Since the magnitude of the actual momentum is smaller than the uncertainty, the value cannot be defined.
Q.62: The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energy lists:
 n = 4, l = 2, ml = –2 , m_{s} = –1/2
 n = 3, l = 2, ml= 1 , m_{s} = +1/2
 n = 4, l = 1, ml = 0 , m_{s} = +1/2
 n = 3, l = 2, ml = –2 , m_{s} = –1/2
 n = 3, l = 1, ml = –1 , m_{s}= +1/2
 n = 4, l = 1, ml = 0 , m_{s} = +1/2
Ans.
For n = 4 and l = 2, the orbital occupied is 4d.
For n = 3 and l = 2, the orbital occupied is 3d.
For n = 4 and l = 1, the orbital occupied is 4p.
Hence, the six electrons i.e., 1, 2, 3, 4, 5, and 6 are present in the 4d, 3d, 4p, 3d, 3p, and 4p orbitals respectively.
Therefore, the increasing order of energies is 5(3p) < 2(3d) = 4(3d) < 3(4p) = 6(4p) < 1 (4d).
Q.63. The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electrons in 4p orbital. Which of these electron experiences the lowest effective nuclear charge?
Ans.
Nuclear charge experienced by an electron (present in a multielectron atom) is dependant upon the distance between the nucleus and the orbital, in which the electron is present. As the distance increases, the effective nuclear charge also decreases. Among porbitals, 4p orbitals are farthest from the nucleus of bromine atom with (+35) charge. Hence, the electrons in the 4p orbital will experience the lowest effective nuclear charge. These electrons are shielded by electrons present in the 2p and 3p orbitals along with the sorbitals. Therefore, they will experience the lowest nuclear charge.
Q.64. Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge?
(i) 2s and 3s,
(ii) 4d and 4f,
(iii) 3d and 3p
Ans.
Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of a multielectron atom. The closer the orbital, the greater is the nuclear charge experienced by the electron (s) in it.
(i)The electron(s) present in the 2s orbital will experience greater nuclear charge (being closer to the nucleus) than the electron(s) in the 3s orbital.
(ii)4d will experience greater nuclear charge than 4f since 4d is closer to the nucleus.
(iii)3p will experience greater nuclear charge since it is closer to the nucleus than 3f.
Q.65.The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus?
Ans.
Nuclear charge is defined as the net positive charge experienced by an electron in a multielectron atom. The higher the atomic number, the higher is the nuclear charge. Silicon has 14 protons while aluminium has 13 protons. Hence, silicon has a larger nuclear charge of (+14) than aluminium, which has a nuclear charge of (+13). Thus, the electrons in the 3p orbital of silicon will experience a more effective nuclear charge than aluminium.
Q.66. Indicate the number of unpaired electrons in:
(a)P
(b)Si
(c)Cr
(d)Fe
(e)Kr
Ans.
(a)Phosphorus (P):
Atomic number = 15
The electronic configuration of P is:
1s 2 2s 2 2p 6 3s 2 3p 3
The orbital picture of P can be represented as:
From the orbital picture, phosphorus has three unpaired electrons.
(b) Silicon (Si):
Atomic number = 14
The electronic configuration of Si is:
1s 2 2s 2 2p 6 3s 2 3p 2
The orbital picture of Si can be represented as:
From the orbital picture, silicon has two unpaired electrons.
(c) Chromium (Cr):
Atomic number = 24
The electronic configuration of Cr is:
1s 2 2s2 2p 6 3s 2 3p 6 4s 1 3d 5
The orbital picture of chromium is:
From the orbital picture, chromium has six unpaired electrons.
(d) Iron (Fe):
Atomic number = 26
The electronic configuration is:
1s 2 2s2 2p 6 3s 2 3p 6 4s 2 3d 6
The orbital picture of chromium is:
From the orbital picture, iron has four unpaired electrons.
(e) Krypton (Kr):
Atomic number = 36
The electronic configuration is:
1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d10 4p 6
The orbital picture of krypton is:
Since all orbitals are fully occupied, there are no unpaired electrons in krypton.
Q.67.
(a) How many subshells are associated with n = 4?
(b) How many electrons will be present in the subshells having ms value of –1/2 for n = 4?
Ans.
(a)n = 4 (Given)
For a given value of ‘n’, ‘l’ can have values from zero to (n – 1).
Therefore, = 0, 1, 2, 3
Thus, four subshells are associated with n = 4, which are s, p, d and f.
(b) Number of orbitals in the n th shell = n^{2}
For n = 4
Number of orbitals = 16
If each orbital is taken fully, then it will have 1 electron with ms value of .
Therefore, Number of electrons with m_{s} value of (\(\frac{1}{2}\)) is 16.
Also Access 
NCERT Exemplar for class 11 chemistry Chapter 2 
CBSE Notes for class 11 chemistry Chapter 2 
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