NCERT Solutions for Class 11 Chemistry Chapter 2 Structure of Atom

NCERT Solutions Class 11 Chemistry Chapter 2 – Free PDF Download

NCERT Solutions for Class 11 Chemistry Chapter 2 Structure of Atom are provided on this page as a free source of educational content for Class 11 students. These NCERT Solutions aim to provide students with comprehensive answers to all questions asked in Chapter 2 of the NCERT Class 11 Chemistry textbook. Chapter 2 will completely explain the structure of an atom. The main aim of creating NCERT Solutions is to help students focus on important concepts and learn them effectively.

Students will face problems in answering the numericals that would appear in the board exam. For this purpose, the faculty at BYJU’S have designed the NCERT Solutions for Class 11 Chemistry Chapter 2 in a stepwise manner, based on the marks weightage allotted by the CBSE for the 2022-23 syllabus. The solutions also contain explanations for each and every step to provide a quality learning experience for the Class 11 students. The NCERT Solutions for Class 11 Chemistry provided on this page can also be downloaded as a PDF, for free, by clicking the download button provided below.

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NCERT Solutions for Class 11 Chemistry Chapter 2 Structure of Atom

Structure of Atom” is the second chapter in the NCERT Class 11 Chemistry textbook. The topics covered under this chapter include subatomic particles, Thomson’s atomic model, Rutherford’s atomic model, Bohr’s model and the quantum mechanical model of the atom.  The types of questions asked in the NCERT exercise section for this chapter include:

  • Basic calculations regarding subatomic particles (protons, electrons and neutrons).
  • Numericals based on the relationship between wavelength and frequency.
  • Numericals based on calculating the energy associated with electromagnetic radiation.
  • Electron transitions to different shells.
  • Writing electron configurations.
  • Questions related to quantum numbers and their combinations (for electrons).

Students can note that these NCERT Solutions have been prepared and solved by our experienced subject experts, as per the latest term-wise CBSE Syllabus 2022-23 and its guidelines. The NCERT Solutions for Class 11 Chemistry provided on this page (for Chapter 2) provide detailed explanations with the steps to be followed while solving the numerical value questions that are frequently asked in term-wise examinations. The subtopics covered in the chapter are listed below.

Subtopics of NCERT Solutions for Class 11 Chemistry Chapter 2 Structure Of Atom

  1. Sub-atomic Particles
    • Discovery Of Electron
    • Charge To Mass Ratio Of Electron
    • Charge On The Electron
    • Discovery Of Protons And Neutrons
  2. Atomic Models
    • Thomson Model Of Atom
    • Rutherford’s Nuclear Model Of Atom
    • Atomic Number And Mass Number
    • Isobars And Isotopes
    • Drawbacks Of Rutherford Model
  3. 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
  4. Bohr’s Model For Hydrogen Atom
    • Explanation Of Line Spectrum Of Hydrogen
    • Limitations Of Bohr’s Model
  5. Towards Quantum Mechanical Model Of The Atom
    • Dual Behaviour Of Matter
    • Heisenberg’s Uncertainty Principle
  6. 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.

Students can make use of the NCERT Solutions for Class 11 exclusively to comprehend key topics and concepts, along with precise answers to exercise questions in the NCERT Class 11 Chemistry Textbook.

Access the answers of NCERT Solutions for Class 11 Chemistry Chapter 2


 

Q.1. (i) Calculate the number of electrons which will together weigh one gram.
(ii) Calculate the mass and charge of one mole of electrons.

Ans:

(i)  Mass of 1 electron = 9.108 x 10-28 g

Hence, 

1 g = 1/(9.108 x 10-28) = 1.098 x 1027 electrons

(ii) Mass of one mole of electron = 9.108 x 10-28 x 6.022 x 1023

We get,

= 5.48 x 10-4 g

Charge on one mole of electron = 1.6 x 10-19 C x 6.022 x 1023

We get,

= 9.63 x 104 C

 

Q.2. (i) Calculate the total number of electrons present in one mole of methane.
(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of 14C.
(Assume that mass of a neutron = 1.675 × 10–27 kg).
(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH3
at STP.
Will the answer change if the temperature and pressure are changed?

Ans:

(i) 1 molecule of methane contains 10 electrons (6 from carbon, 4 from hydrogen)

Therefore, 1 mole of methane contains = 10 x 6.022 x 1023 = 6.022 x 1024 electrons.

(ii) (a)1 g atom of 14C = 14 g 

Since 14 g = 14000 mg have 6.022 x 1023 x 8 neutrons

Therefore, 7 mg will have neutrons = (6.022 x 1023 x 8)/14000 x 7 = 2.4088 x 1022

(b) mass of 1 neutron = 1.675 x 10-27kg

Hence, mass of 2.4088 x 1021 neutrons = 2.4088 x 1021 x 1.67 x 10-27 = 4.0347 x 10-6 kg

(iii) Step I. Calculation of total number of NHmolecules

Gram molecular mass of ammonia (NH3) = 17 g = 17 x 103 mg

17 x 103  mg of NH3 have molecules = 6.022 x 1023

34 mg of NH3 have molecules = 6.022×1023(17×103mg)×(34mg)=1.2044×1020molecules\frac{6.022 \times 10^{23}}{(17\times 10^{3}mg)}\times (34mg)\\= 1.2044 \times 10^{20}molecules

Step II. Calculation of total number and mass of protons

No. of protons present in one molecule of NH3 = 7 + 3 = 10

No. of protons present in 12.044 x 1020 molecules of NH3 = 12.044 x 1020 x 10

= 1.2044 x 1022 protons

Mass of one proton = 1.67 x 10-27 kg

Mass of 1.2044 x 1022 Protons = (1.67 x 10-27 kg) x 1.2044 x 1022

= 2.01 x 10-5 kg

No, the answer will not change upon changing the temperature and pressure because only the number of protons and mass of protons are involved.

 

Q.3. How many neutrons and protons are there in the following nuclei?

613C,816O,1224Mg,2656Fe,3888Sr_{6}^{13}\textrm{C}\: ,\: _{8}^{16}\textrm{O}\: ,\: _{12}^{24}\textrm{Mg}\: ,\: _{26}^{56}\textrm{Fe}\: ,\: _{38}^{88}\textrm{Sr}

Ans:

613C_{6}^{13}\textrm{C}:

Mass number = 13

Atomic number = Number of protons = 6

Therefore, total number of neutrons in 1 carbon atom = Mass number – Atomic number = 13 – 6 = 7

816O_{8}^{16}\textrm{O} :

Mass number = 16

Atomic number = Number of protons = 8

Therefore, number of neutrons = Mass number – Atomic number = 16 – 8 = 8

1224Mg _{12}^{24}\textrm{Mg} :

Mass number = 24

Atomic number = Number of protons = 12

Number of neutrons = Mass number – Atomic number = 24 – 12 = 12

2656Fe _{26}^{56}\textrm{Fe} :

Mass number = 56

Atomic number = Number of protons = 26

Number of neutrons = Mass number – Atomic number = 56 – 26 = 30

3888Sr _{38}^{88}\textrm{Sr} :

Mass number = 88

Atomic number = Number of protons = 38

Number of neutrons = Mass number – Atomic number = 88 – 38 = 50

 

Q.4. Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A)

(I)Z = 17, A = 35

(II)Z = 92, A = 233

(III)Z = 4, A = 9

Ans:

(I) The atom with atomic number 17 and mass number 35 is chlorine. Therefore, its complete symbol is 1735Cl_{17}^{35}\textrm{Cl}

(II) The atom with atomic number 92 and mass number 233 is uranium. Therefore, its complete symbol is 92233U_{92}^{233}\textrm{U}

(III) The atom with atomic number 4 and mass number 9 is beryllium. Therefore, its complete symbol is 49Be_{4}^{9}\textrm{Be}

Q.5. Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the frequency (ν) and wavenumber (ν ) of the yellow light.

Ans: Rearranging the expression,

λ=cν\lambda =\frac{c}{\nu }

the following expression can be obtained,

ν=cλ\nu =\frac{c}{ \lambda }      ……….(1)

Here, ν\nu denotes the frequency of the yellow light

c denotes the speed of light ( 3×108m/s3\times 10^{8}\, m/s )

λ\lambda denotes the wavelength of the yellow light (580 nm, 580×109m/s580\times 10^{-9}\, m/s )

Substituting these values in eq. (1):

ν=3×108580×109=5.17×1014s1\nu =\frac{3\times 10^{8}}{580\times 10^{-9}}=5.17\times 10^{14}\, s^{-1}

Therefore, the frequency of the yellow light which is emitted by the sodium lamp is:

5.17×1014s15.17\times 10^{14}\, s^{-1}

The wave number of the yellow light is νˉ=1λ\bar{\nu }=\frac{1}{\lambda } =1580×109=1.72×106m1=\frac{1}{580\times 10^{-9}}=1.72\times 10^{6}\, m^{-1}

 

Q.6. Find the energy of each of the photons which
(i) correspond to light of frequency 3×1015 Hz.
(ii) have a wavelength of 0.50 Å.

Ans:

(i)

The energy of a photon (E) can be calculated by using the following expression:

E= hνh\nu

Where, ‘h’ denotes Planck’s constant, which is equal to 6.626×1034Jsν6.626\times 10^{-34}\, Js\\\nu (frequency of the light) = 3×10153\times 10^{15}Hz

Substituting these values in the expression for the energy of a photon, E:

E=(6.626×1034)(3×1015)E=1.988×1018JE=(6.626\times 10^{-34})(3\times 10^{15})\\ \\ E=1.988\times 10^{-18}\, J

(ii)

The energy (E) of a photon having wavelength (λ) is given by the expression:

E=hc/λ

Where,

h (Planck’s constant) = 6.626×1034Js6.626\times 10^{-34}Js

c (speed of light) = 3×108m/s3\times 10^{8}\, m/s

Substituting these values in the equation of ‘E’:

E=(6.626×1034)(3×108)0.50×1010=3.976×1015JE=3.98×1015JE=\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 wavenumber of a light wave whose period is 2.0 × 10–10 s.

Ans: Frequency of the light wave (ν\nu) = 1Period1period=12.0×1010s=5.0×109s1\frac{1}{Period}\\\frac{1}{period}=\frac{1}{2.0 \times 10^{-10}s}\\=5.0\times 10^{9}s^{-1}

Wavelength of the light wave(λ\lambda) =cν\frac{c}{\nu}

Where,

c denotes the speed of light,  3×108m/s3\times 10^{8}\, m/s

Substituting the values in the given expression of λ\lambda:

λ=3×1085.0×109=6.0×102m\lambda =\frac{3\times 10^{8}}{5.0\times 10^{9}}=6.0\times 10^{-2}m

Wave number (νˉ)(\bar{\nu }) of light = 1λ=16.0×102=1.66×101m1=16.66m1\frac{1}{\lambda }=\frac{1}{6.0\times 10^{-2}}=1.66\times 10^{1}\, m^{-1}=16.66\, m^{-1}

 

Q.8. What is the number of photons of light with a wavelength of 4000 pm that provides 1J of energy?

Ans: Energy  of photon (E) = hcλ\frac{hc}{\lambda }

Where, λ\lambda is the wavelength of the photons = 4000×1012m=4×109m4000\times 10^{-12}\, m = 4 \times 10^{-9}m

c denotes the speed of light in vacuum =3×108m/s3\times 10^{8}\, m/s

h is Planck’s constant, whose value is 6.626×1034Js6.626\times 10^{-34}\, Js

Substituting these values, we get,

E=(6.626×1034Js)×(3×108ms1)(4×109m)E = \frac{(6.626 \times 10^{-34}Js)\times (3\times 10^{8}ms^{-1})}{(4\times 10^{-9}m)}

= 4.969 x 10-17 J

Hence, 1J energy of photons = 14.969×1017=2.012×1016photons\frac{1}{4.969\times 10^{-17}} = 2.012 \times 10^{16}photons

 

Q.9. A photon of wavelength 4 × 10–7 m strikes on metal surface, the work function of the metal is 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 × 10–19 J).

Ans:

(i)

Energy of the photon (E)= hν=hcλh\nu =\frac{hc}{\lambda }

Where, h denotes Planck’s constant, whose value is 6.626×1034Js6.626\times 10^{-34}\,Js

c denotes the speed of light = 3×108m/sλ3\times 10^{8}\,m/s\\\lambda= wavelength of the photon =4×107m/s4\times 10^{-7}\,m/s

Substituting these values, we get,

E=(6.626×1034)(3×108)4×107=4.9695×1019JE=\frac{(6.626\times 10^{-34})(3\times 10^{8})}{4\times 10^{-7}}=4.9695\times 10^{-19}\, J

Therefore, energy of the photon = 4.97×1019J4.97\times 10^{-19}\, J

(ii)

The kinetic energy of the emission EkE_{k} is given by

=hνhν0=(EW)eV=(4.9695×10191.6020×1019)eV2.13eV=(3.10202.13)eV=0.9720eV=h\nu -h\nu _{0}\\ \\ =(E-W)eV\\ \\ =(\frac{4.9695\times 10^{-19}}{1.6020\times 10^{-19}})eV-2.13\, eV\\ \\ =(3.1020-2.13)eV\\ \\ =0.9720\, eV

Therefore, the kinetic energy of the emission = 0.97 eV.

(iii)

The velocity of the photoelectron (v) can be determined using the following expression:

12mv2=hνhν0v=2(hνhν0)m\frac{1}{2}mv^{2} =h\nu -h\nu _{0}\\ \\ \Rightarrow v =\sqrt{\frac{2(h\nu -h\nu _{0})}{m}}

Where (hνhν0)(h\nu -h\nu _{0}) is the K.E of the emission (in Joules) and ‘m’ denotes the mass of the photoelectron.

Substituting the values in the given expression of  v:

v=2×(0.9720×1.6020×1019)J9.10939×1031kg=0.3418×1012m2s2v=5.84×105ms1v=\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}

Therefore, the velocity of the  photoelectron is 5.84×105ms15.84\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: Ionization energy (E) of sodium = NAhcλ=(6.023×1023mol1)(6.626×1034)Js(3×108)ms1242×109m=4.947×105Jmol1=494.7×103Jmol1=494kJmol1\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 emits monochromatic yellow light of the wavelength of 0.57µm. Calculate the rate of emission of quanta per second.

Ans: Power of the bulb, P = 25 Watt = 25Js125\, Js^{-1}

Energy (E) of one photon = hν=hcλh\nu = \frac{hc}{\lambda }

Substituting these values in the expression of E:

E=(6.626×1034)(3×108)(0.57×106)=34.87×1020JE=\frac{(6.626\times 10^{-34})(3\times 10^{8})}{(0.57\times 10^{-6})}=34.87\times 10^{-20}\, J

E= 34.87×1020J34.87\times 10^{-20}\, J

Thus, the rate of emission of quanta (per second) = E=2534.87×1020=7.169×1019s1E=\frac{25}{34.87\times 10^{-20}}=7.169\times 10^{19}\, s^{-1}

Q.12. Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (ν0 ) and work function (W0 ) of the metal.

Ans: Wavelength of the radiation, λ\lambda = 6800 Å=6800×1010m6800\times 10^{-10}\, m

Threshold frequency of the metal (ν0\nu _{0} ) =cλ0=3×108ms16.8×107m=4.41×1014s1\frac{c}{\lambda _{0}}=\frac{3\times 10^{8}ms^{-1}}{6.8\times 10^{-7}m}=4.41\times 10^{14}\, s^{-1}

Therefore, work function, (W0) = hν0=(6.626×1034Js)(4.41×1014s1)=2.922×1019Jh\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 emitted when the electron in a hydrogen atom undergoes the transition from an energy level with n = 4 to an energy level with n = 2?

Ans: The nin_{i} = 4 to nfn_{f} = 2 transition results in a spectral line of the Balmer series. The energy involved in this transition can be calculated using the following expression:

E=2.18×1018[1ni21nf2]E=2.18\times 10^{-18}[\frac{1}{n_{i}^{2}}-\frac{1}{n_{f}^{2}}]

Substituting these values in the expression of E:

E=2.18×1018[142122]=2.18×1018[1416]=2.18×1018×(316)E=(4.0875×1019J)E=2.18\times 10^{-18}[\frac{1}{4^{2}}-\frac{1}{2^{2}}]\\ \\ =2.18\times 10^{-18}[\frac{1-4}{16}]\\ \\ =2.18\times 10^{-18}\times (-\frac{3}{16})\\ \\ E=-(4.0875\times 10^{-19}J)

Here, the -ve sign denotes the emitted energy.

Wavelength of the emitted light (λ)=hcE(\lambda )=\frac{hc}{E}

Since E = hc/λ

Substituting these values in the expression of λ

λ=(6.626×1034)(3×108)(4.0875×1019)=4.8631×107mλ=486.31×109m=486nm\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. How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H atom (energy required to remove the electron from n =1 orbit).

Ans: The expression for the ionization energy is given by,

En=(2.18×1018)Z2n2E_{n} =\frac{-(2.18\times 10^{-18})Z^{2}}{n^{2}}

Where Z denotes the atomic number and n is the principal quantum number

For the ionization from n1=5n_{1}=5 to n2=n_{2} =∞,

ΔE=EE5=[((2.18×1018J)(1)2()2)((2.18×1018J)(1)2(5)2)]=0.0872×1018JΔE=8.72×1020J\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

Therefore, the required energy for the ionization of hydrogen from n = 5 to n = ∞  is 8.72×1020J8.72\times 10^{-20}\, J

Energy required for n1 = 1 to n = ∞

[{(2.18×1018J)(12)()2}{(2.18×1018J)(12)(12)}]\left [\left\{\frac{-\left ( 2.18\times 10^{-18}J \right )\left ( 1^{2} \right )}{\left ( \infty \right )^{2}} \right\} -\left\{-\frac{\left ( 2.18 \times 10^{-18}J \right )(1^{2})}{(1^{2})} \right\} \right ]

= 2.18 x 10-18 J

ΔE=2.18×1018J\Delta E = 2.18 \times 10^{-18}J

Hence, a lower amount of energy is required in order to ionize electrons in the 5th orbital of a hydrogen atom when compared to that in the ground state.

Q.15. What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state?

Ans. A total number of  15 lines (5 + 4 + 3 + 2 + 1) will be obtained in this hydrogen emission spectrum.

The number of spectral lines emitted when an electron in the nth level drops down to the ground state is given by

n(n1)2\frac{n(n-1)}{2}

Since n = 6, total no. spectral lines = 6(61)2=15\frac{6(6-1)}{2}=15

Hence, the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state is 15

Q.16. (i) The energy associated with the first orbit in the hydrogen atom is –2.18 × 10–18 J atom–1. What is the energy associated with the fifth orbit? (ii) Calculate the radius of Bohr’s fifth orbit for the hydrogen atom.

Ans. The expression for the energy associated with nth orbit in hydrogen atom is

En = (-2.18 x 10-18)/n2 J/atom

(i) Energy associated with the fifth orbit of hydrogen atom is calculated as:

E5=(2.18×1018)(5)2=(2.18×1018)25=8.72×1020Jatom1E_{5}=\frac{-(2.18\times 10^{-18})}{(5)^{2}}=\frac{-(2.18\times 10^{-18})}{25}=-8.72\times 10^{-20}\, J atom^{-1}

(ii) Radius of Bohr’s nth orbit for hydrogen atom is given by, rn = (0.0529 nm) n 2

For, n = 5

r5=(0.0529nm)(52)r5=1.3225nmr_{5}=(0.0529\, nm)(5^{2})\\ \\ r_{5}=1.3225\, nm

Q.17. Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.

Ans. For the Balmer series of the hydrogen emission spectrum, ni = 2. Therefore, the expression for the wavenumber(νˉ\bar{\nu}) is:

νˉ=[1(2)21nf2](1.097×107m1)\bar{\nu}=[\frac{1}{(2)^{2}}-\frac{1}{n_{f}^{2}}](1.097\times 10^{7}\, m^{-1})

Since wave number(νˉ\bar{\nu}) is inversely proportional to the transition wavelength, the lowest possible value of (νˉ\bar{\nu}) corresponds to the longest wavelength transition.

For (νˉ\bar{\nu}) to be of the lowest possible value, nf should be minimum. In the Balmer series, transitions from ni = 2 to nf = 3 are allowed.

Hence, taking nf = 3, we get:

νˉ=(1.097×107)[1(2)2132]νˉ=(1.097×107)[1419]=(1.097×107)[9436]=(1.097×107)[536]νˉ=1.5236×106m1\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{9-4}{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. Step I. Calculation of energy required

The energy of electron (En) = 2.18×1011n2ergs=2.18×1018Jn2(1J=107ergs)\frac{2.18\times 10^{-11}}{n^{2}}ergs = \frac{-2.18\times 10^{-18}J}{n^{2}} (\because 1J = 10^{7}ergs)

The energy in Bohr’s first orbit (E1) = 2.18×1018J(1)2=2.18×1018J1\frac{-2.18\times 10^{-18}J}{(1)^{2}}= \frac{-2.18 \times 10^{-18}J}{1}

The energy in Bohr’s fifth orbit (E5) = 2.18×1018J(5)2=2.18×1018J25\frac{-2.18\times 10^{-18}J}{(5)^{2}}= \frac{-2.18 \times 10^{-18}J}{25}

∴ Energy required (ΔE)=E5E1=(2.1825×1018J)(2.181×1018J)=2.18×1018(1125)J=2.18×1018×2425=2.09×1018J(\Delta E)= E_{5}- E_{1}= (-\frac{2.18}{25}\times 10^{-18}J)- (-\frac{2.18}{1}\times 10^{-18}J)\\=2.18 \times 10^{-18}(1-\frac{1}{25})J\\= 2.18\times 10^{-18}\times \frac{24}{25}= 2.09 \times 10^{-18}J

Step II. Calculation of wavelength of light emitted

ΔE=hν=hcλλ=hcΔE=(6.626×1034Js)×(3×108ms1)(2.09×1018J)=9.50×108m\Delta E= h\nu = \frac{hc}{\lambda }\\\therefore \lambda =\frac{hc}{\Delta E}= \frac{(6.626\times 10^{-34}J-s)\times (3\times 10^{8}ms^{-1})}{(2.09\times 10^{-18}J)}\\= 9.50 \times 10^{-8}m

= 950 Å  (∵ 1Å = 10-10m)

 

Q.19. The electron energy in hydrogen atom is given by En= (–2.18 × 10–18 )/n2 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. En=(2.18×1018)(n)2JE_{n}=\frac{-(2.18\times 10^{-18}) }{(n)^{2} }J

Required energy for the ionization from n = 2 is:

ΔE=EE2=[((2.18×1018)()2)((2.18×1018)(2)2)]J=[(2.18×1018)40]J=0.545×1018JΔE=5.45×1019J\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

Now we know that,

λ=hcΔE\lambda =\frac{hc}{\Delta E}

Here, λ is the longest wavelength causing the transition

E=(6.626×1034)(3×108)(5.45×1019)=3.647×107m=3647×1010E=\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 × 107 m s–1

Ans. As per de Broglie’s equation,

λ=hmv\lambda =\frac{h}{mv }

Where, λ denotes the wavelength of the moving particle

m is the mass of the particle

v denotes the velocity of the particle

h is Planck’s constant

Substituting these values we get,

λ=(6.626×1034)Js(9.10939×1031kg)(2.05×107ms1)=3.548×1011m\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

Therefore, the wavelength of an electron moving with a velocity of 2.05×107ms1is3.548×1011m2.05\times 10^{7}\, ms^{-1}\: is\:\: 3.548\times 10^{-11}\, m

Q.21. The mass of an electron is 9.1 × 10–31 kg. If its K.E. is 3.0 × 10–25 J, calculate its wavelength.

Ans.

The wavelength of an electron can be found by de Broglie’s equation:

λ=hmν\lambda = \frac{h}{m\nu }

Given the K.E. of electron 3.0 x 10-25 J which is equal to 12mν2\frac{1}{2}m\nu ^{2}

Hence we get,

12mν2=3.0×1025Jν=2KEm=2(3.0×1025J)9.1×1031kg=811.579m/s\frac{1}{2}m\nu ^{2}= 3.0\times 10^{-25}J\\\nu = \sqrt{\frac{2KE}{m}}= \sqrt{\frac{2(3.0\times 10^{-25}J)}{9.1\times 10^{-31}kg}} = 811.579 m/s

Hence the wavelength is given by,

λ=6.626×1034Js(9.1×1031kg)(811.579m/s)=8.9625×107m\lambda = \frac{6.626\times 10^{-34}Js}{(9.1\times 10^{-31}kg)(811.579m/s)}\\= 8.9625\times 10^{-7}m

 

Q. 22. Which of the following are isoelectronic species i.e., those having the same number
of electrons?
Na+ , K+ , Mg2+, Ca2+, S2–, Ar.

Ans:

Number of electrons present in different species are:

11Na+ = 11 – 1 = 10;             19K+ = 19 – 1 = 18;

12Mg2+ = 12 – 2 = 10;

20Ca2+ = 20 – 2 = 18;         16S2- = 16 + 2 = 18

Ar = 18

Hence isoelectronic species are

(i) Naand Mg2+

(having number of electrons = 10)

(ii) K+, Ca2+, S2- and Ar (having number of electrons = 18)

 

Q.23. (I)Write the electronic configurations of the following ions:

(a)H

(b)Na+

(c)O2–

(d)F –

(II) What are the atomic numbers of elements whose outermost electrons are represented by

(a) 3s1

(b) 2p3 and

(c) 3p5?

(III) Which atoms are indicated by the following configurations?

(a)[He] 2s1

(b)[Ne] 3s2 3p3

(c)[Ar] 4s2 3d1 .

Ans:

(I) (a) H  ion

The electronic configuration of the Hydrogen atom (in its ground state) 1s1. The single negative charge on this atom indicates that it has gained an electron. Thus, the electronic configuration of H = 1s2

(b) Naion

Electronic configuration of Na =  1s2 2s2p6 3s1 . Here, the +ve charge indicates the loss of an electron.

∴Electronic configuration of Na+ = 1s2 2s2 2p6

(c) O2– ion

Electronic configuration of 0xygen = 1s 2 2s 2 2p 4. The ‘-2 ‘ charge suggests that it has gained 2 electrons.

∴ Electronic configuration of O2– ion = 1s 2 2s 2 2p 6

(d) F – ion

Electronic configuration of Fluorine = 1s22s22p5. The negative charge indicates that it has gained one electron

∴ Electronic configuration of F ion = 1s 2 2s 2 2p 6

(II) (a) 3s1

Complete electronic configuration: 1s 2 2s 2 2p 6 3s 1.

Total number of electrons in the atom = 2 + 2 + 6 + 1 = 11

∴ The element’s atomic number is 11

(b) 2p 3

Complete electronic configuration: 1s 2 2s 2 2p 3 .

Total number of electrons in the atom =2 + 2 + 3 = 7

∴The element’s atomic number is 7

(c) 3p 5

Complete electronic configuration: 1s 2 2s 2 2p 6 3s2 3p5 .

Total number of electrons in the atom = 2 + 2 + 6 + 2 + 5 = 17

∴ The element’s atomic number is 17

(III)(a)[He] 2s 1

Complete electronic configuration: 1s 2 2s 1 .

∴ The element’s atomic number is 3 . The element is lithium (Li)

(b)[Ne] 3s 2 3p3

Complete electronic configuration: 1s 2s 2 2p 6 3s 3p 3 .

∴ The element’s atomic number is 15. The element is phosphorus (P).

(c)[Ar] 4s 2 3d1

Complete electronic configuration: 1s 2 2s 2p 6 3s 2 3p 6 4s 3d 1 .

∴ The element’s atomic number is 21. The element is scandium (Sc).

 

Q.24. What is the lowest value of n that allows g orbitals to exist?

Ans. For g-orbitals, l = 4.

For any given value of ‘n’, the possible values of ‘l’ range from 0 to (n-1).

∴ For l = 4 (g orbital), least value of n = 5.

 

Q.25. An electron is in one of the 3d orbitals. Give the possible values of n, l and ml for this electron.

Ans: For the 3d orbital:

Possible values of the Principal quantum number (n) = 3

Possible values of the Azimuthal quantum number (l) = 2

Possible values of the Magnetic quantum number (ml) = – 2, – 1, 0, 1, 2

 

Q.26. 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)

In a neutral atom, number of protons = number of electrons.

∴ Number of protons present in the atoms of the element = 29

(ii)

The electronic configuration of this element (atomic number 29) is 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 10 . The element is copper (Cu).

 

Q.27.Give the number of electrons in the species , H2+, H2 and O2+

Ans: No. electrons present in H2 = 1 + 1 = 2.

∴ Number of electrons in H2+ = 2 – 1 = 1

Number of electrons in H2 = 1 + 1 = 2

Number of electrons in O2 = 8 + 8 = 16.

∴ Number of electrons in O2+= 16 – 1 = 15

 

Q.28. (I)An atomic orbital has n = 3. What are the possible values of l and ml ?

(II)List the quantum numbers (mand l) of electrons for 3d orbital.

(III) Which of the following orbitals are possible? 1p, 2s, 2p and 3f

Ans.

(I)

The possible values of ‘l’ range from 0 to (n – 1). Thus, for n = 3, the possible values of l are 0, 1, and 2.

The total number of possible values for ml = (2l + 1). Its values range from -l to l.

For n = 3 and l = 0, 1, 2:

m= 0

m1 = – 1, 0, 1

m2 = – 2, – 1, 0, 1, 2

(II)

For 3d orbitals, n = 3 and  l = 2. For l = 2 , possible values of m= –2, –1, 0, 1, 2

(III)

It is possible for the 2s and 2p orbitals to exist. The 1p and 3f cannot exist.

For the 1p orbital, n=1 and l=1, which is not possible since the value of l must always be lower than that of n.

Similarly, for the 3f orbital, n =3 and l = 3, which is not possible.

Q.29. Using s, p and 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 implies a 1s orbital.

(b)n = 3 and l = 1 implies a 3p orbital.

(c)n = 4 and l = 2 implies a 4d orbital.

(d)n = 4 and l = 3 implies a 4f orbital.

Q.30. Explain, giving reasons, which of the following sets of quantum numbers are not possible.

a) n = 0, l = 0, ml= 0, m=+12\frac{1}{2}

b) n = 1, l = 0, ml= 0, m=-12\frac{1}{2}

c) n = 1, l = 1, ml= 0, m=+12\frac{1}{2}

d) n = 2, l = 1, ml= 0, m=-12\frac{1}{2}

e) n = 3, l = 3, ml= -3, m=+12\frac{1}{2}

f) n = 3, l = 1, ml= 0, m=+12\frac{1}{2}

Ans. (a) Not possible. The value of n cannot be 0.

(b) Possible.

(c) Not possible. The value of l cannot be equal to that of n.

(d) Possible.

(e) Not possible. For n = 3, l cannot be 3

(f) Possible.

Q.31. How many electrons in an atom may have the following quantum numbers?

a) n = 4, m=-12\frac{1}{2}

b)n = 3, l = 0

Ans.(a)If n is the principal quantum number, the total number of electrons in the atom = 2n 2

For n = 4, Total number of electrons = 2 (4)2 = 32 and half of them have ms = -1/2 Number of electrons (having n = 4 and m=-12\frac{1}{2}) = 16

(b)n = 3, l = 0 indicates the 3s orbital. Therefore, number of electrons with n = 3 and l = 0 is 2.

Q.32. 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. Hydrogen atom have only one electron. As per Bohr’s postulates, the angular momentum of this electron is:

mvr=nh2π.(1)mvr=n\frac{h}{2\pi }\: \: \: ….(1)

Where, n = 1, 2, 3, …

As per de Broglie’s equation:

λ=hmvormv=hλ.(2)\lambda =\frac{h}{mv }\\ \\ or\: mv=\frac{h}{\lambda }\: \: \: \: …….(2)

Substituting the value of mv from equation (2) in equation (1) we get,

hrλ=nh2πor2πr=nλ..(3)\frac{hr}{\lambda }=n\frac{h}{2\pi }\\ \\ or\: 2\pi r=n\lambda \: \: \: \: …..(3)

But ‘2πr’ is the Bohr orbit’s circumference. Therefore, equation (3) proves that the Bohr orbit’s circumference for the hydrogen atom is an integral multiple of the de Broglie wavelength of the electron, which is revolving around the orbit.

Q.33. What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of Hespectrum?

Ans. The wave number associated with the Balmer transition for the He+ ion (n = 4 to n = 2 ) is given by:

νˉ=1λ=RZ2(1n121n22)\bar{\nu}=\frac{1}{\lambda }=RZ^{2}(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}})

Where, n1 = 2 n2 = 4

Z = atomic number of helium

νˉ=1λ=R(2)2(14116)=4R(4116)νˉ=1λ=3R4λ=43R\bar{\nu}=\frac{1}{\lambda }=R(2)^{2}(\frac{1}{4}-\frac{1}{16})\\ \\ =4R(\frac{4-1}{16})\\ \\ \bar{\nu}=\frac{1}{\lambda }=\frac{3R}{4}\\ \\ \Rightarrow \lambda =\frac{4}{3R}

As per the question, the desired transition in the hydrogen spectrum must have the same wavelength as that of Hespectrum.

R(1)2[1n121n22]=3R4[1n121n22]=34.(1)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)

This equality is true only when the value of n1 = 1 and n2 = 2.

Hence, the transition for n2 = 2 to n = 1 in the hydrogen spectrum would have the same wavelength as the Balmer transition from n = 4 to n = 2 of the He+ spectrum.

Q.34. Calculate the energy required for the process

He(g)+He(g)2++eHe^{+}_{(g)}\rightarrow He^{2+}_{(g)}+e^{-}

The ionization energy for the H atom in the ground state is 2.18×1018Jatom12.18\times 10^{-18}\, J\,atom^{-1}

Ans. The energy associated with hydrogen-like species is:

En=2.18×1018(Z2n2)JE_{n}=-2.18\times 10^{-18}(\frac{Z^{2}}{n^{2}})J

For the ground state of the hydrogen atom,

ΔE=EE1=0[(2.18×1018(1)2(1)2]JΔE=2.18×1018J\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)+He(g)2++eHe^{+}_{(g)}\rightarrow He^{2+}_{(g)}+e^{-}

An electron is removed from n = 1 to n = ∞.

ΔE=EE1=0[(2.18×1018(2)2(1)2]JΔE=8.72×1018J\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 required energy for this process is 8.72×1018J8.72\times 10^{-18}J\\.

Q.35. 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 the length of the scale of length 20 cm long.

Ans. 1 m = 100 cm

1 cm = 10–2 m

Length of the scale = 20 cm =20×102m20\times 10^{-2}\, m

Diameter of one carbon atom = 0.15 nm =0.15×109m0.15\times 10^{-9}\, m

Space occupied by one carbon atom 0.15×109m0.15\times 10^{-9}\, m.

Number of carbon atoms that can be placed in a straight line 20×102m0.15×109m=133.33×107=1.33×109\frac{20\times 10^{-2}m}{0.15\times 10^{-9}m}\\ \\ =133.33\times 10^{7}\\ \\ =1.33\times 10^{9}

Q.36. 2×108m2\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 arrangement = 2.4 cm

Number of carbon atoms present = 2 × 108

The diameter of the carbon atom = 2.4×102m2×108m=1.2×1010m\frac{2.4\times 10^{-2}m}{2\times 10^{8}m}\\ \\ =1.2\times 10^{-10}m\\∴Radius of carbon atom=Diameter2=1.2×1010m2=6.0×1011m\frac{Diameter}{2}\\ \\ =\frac{1.2\times 10^{-10}m}{2}\\ \\ =6.0\times 10^{-11}m

 

Q.37. The diameter of the 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) Given 

Diameter of zinc atom = 2.6 Å

= 2.6 x 10-10 m

= 260 x 10-12 m

= 260 pm

Radius of zinc atom in pm = (260 pm)/2 = 130 pm

(b) Given

Length of the arrangement = 1.6 cm

= 1.6 x 10-2 m

Diameter of a zinc atom = 2.6 Å = 2.6 x 10-10 m

∴ Number of zinc atom present in the arrangement

= (1.6 x 10-2 m)/(2.6 x 10-10 m)

= 0.6153 x 108 m

= 6.153 x 107 m

 

Q.38. A certain particle carries 2.5×1016C 2.5 \times 10^{-16}C  of static electric charge. Calculate the number of electrons present in it.

Ans.

Charge carried by one electron = 1.6022×1019C 1.6022 \times 10^{-19}C

Therefore, number of electrons present in a particle carrying 2.5×1016C=11.6022×1019C(2.5×1016C)=1.560×103C=1560C 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, the static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is 1.282×1018C 1.282 \times 10^{-18}C , calculate the number of electrons present on it.

Ans.

Charge on the oil drop =1.282×1018C 1.282 \times 10^{-18}C

Charge on one electron =1.6022×1019C 1.6022 \times 10^{-19}C

Therefore, number of electrons present on the oil drop

1.282×1018C1.6022×1019C=0.8001×101=8.0\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 α-particles. If the thin foil of light atoms like Aluminium etc. is used, what difference would be observed from the above results?

Ans.

Rutherford used alpha rays scattering to reveal the structure of the atom in 1911. The nucleus of heavy atoms is big and carries a lot of positive charge. As a result, when certain alpha particles hit the nucleus, they are easily deflected back. A number of alpha particles are also deflected at tiny angles due to the nucleus’s substantial positive charge. If light atoms are used, their nuclei will be light and their nuclei will have a modest positive charge. As a result, the number of particles deflected back and those deflected at an angle will be insignificant.

 

Q.41. Symbols 3579Brand79Br_{35}^{79}\textrm{Br}\: and\: _{}^{79}\textrm{Br} can be written, whereas symbols 7935Brand35Br_{79}^{35}\textrm{Br}\: and\: _{}^{35}\textrm{Br} are not acceptable. Answer briefly.

Ans.

The general convention followed while representing elements along with their atomic masses (A), and their atomic numbers (Z) is ZAX_{Z}^{A}\textrm{X}