NCERT Solutions for Class 11 Chemistry: Chapter 2 (Structure of Atom)

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 solutions aim to provide students with comprehensive answers to all questions asked in Chapter 2 of the NCERT Class 11 Chemistry textbook. The NCERT Solutions provided in 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 Chemistry – Class 11, 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 CBSE syllabus 2020-21. 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 examinations. The subtopics covered under the chapter are listed below.

Subtopics Of Class 11 Chemistry Chapter 2 Structure Of The 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.

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:

1 electron weighs 9.109*10-31 kg. Therefore, number of electrons that weigh 1 g (10-3 kg) =

10-3kg/9.109*10-31 kg = 1.098*1027 electrons

(ii)

Mass of one mole of electrons = NA* mass of one electron

= (6.022*1023)*(9.109*10-31 kg) = 5.48*10-7 kg

Charge on one mole of electrons = NA* charge of one electron

= (6.022*1023)*(1.6022*10-19 C) = 9.65*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*NA = 6.022*1024 electrons.

(ii) Number of neutrons in 14g (1 mol) of 14C = 8*NA = 4.817*1024 neutrons.

Number of neutrons in 7 mg (0.007g) = (0.007/14)*4.817*1024 = 2.409*1021 neutrons.

Mass of neutrons in 7 mg of 14C = (1.67493*10-27kg)*(2.409*1021) = 4.03*10-6kg

(iii) Molar mass of NH3 = 17g

Number of protons in 1 molecule of NH3 = 7+3 = 10

Therefore, 1 mole (17 grams) of NH3 contains 10*NA = 6.022*1024 protons.

34 mg of NH3 contains (34/1700)*6.022*1024 protons = 1.204*1022 protons.

Total mass accounted for by protons in 34 mg of NH3 = (1.67493*10-27kg)*(1.204*1022) = 2.017*10-5kg.

These values remain constant regardless of any change in temperature and pressure (since these factors do not affect the number of protons in the atom and the mass of each proton).

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

613C,816O,1224Mg,1656Fe,3888Sr_{6}^{13}\textrm{C}\: ,\: _{8}^{16}\textrm{O}\: ,\: _{12}^{24}\textrm{Mg}\: ,\: _{16}^{56}\textrm{Fe}\: ,\: _{38}^{88}\textrm{Sr}

Ans:

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

Mass number of carbon-13 = 13

Atomic number of carbon = Number of protons in one carbon atom = 6

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

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

Mass number of oxygen-16 = 16

Atomic number of oxygen = Number of protons = 8

Therefore, No. neutrons = Mass number – Atomic number = 16 – 8 = 8

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

Mass number = 24

Atomic number = No. protons = 12

No. neutrons = Mass number – Atomic number = 24 – 12 = 12

1656Fe _{16}^{56}\textrm{Fe} :

Mass nubmer = 56

Atomic number of iron = No. protons in iron= 26

No. neutrons = Mass number – Atomic number = 56 – 26 = 30

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

Mass number = 88

Atomic number = No. protons = 38

No. 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)1735C_{17}^{35}\textrm{C}

(II)92233U_{92}^{233}\textrm{U}

(III)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 6.626×1034Js6.626\times 10^{-34}\, Js ν\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 of a photon whose wavelength is (λ\lambda) is:

E= hcνhc\nu

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 for ‘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) = 1Period\frac{1}{Period} 1Period=12.0×1010s=5.0×109s1\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) =cνc\nu

Where,

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

Substituting the value of ‘c’ in the previous expression for 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=16.66m\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 number of photons of light with a wavelength of 4000 pm that provide 1J of energy?

Ans: Energy  of one photon (E) = hνh\nu

Energy of ‘n’ photons (EnE_{n}) = nhνnh\nu n=Enλhc\Rightarrow n=\frac{E_{n}\lambda }{hc}

Where, λ\lambda is the wavelength of the photons = 4000 pm = 4000×1012m4000\times 10^{-12}\, 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 in the expression for n:

n=1×(4000×1012)(6.626×1034)(3×108)=2.012×1016n=\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 2.012×10162.012\times 10^{16}

 

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/s3\times 10^{8}\,m/s λ\lambda= wavelength of the photon =4×107m/s4\times 10^{-7}\,m/s

Substituting these values in the expression for E:

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} can be calculated as follows:

=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.El of the emission (in Joules) and ‘m’ denotes the mass of the photoelectron.

Substituting thes values in the expression for 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 ejected 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=hcνh\nu =\frac{hc}{\nu }

Substituting these values in the expression for 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 discharge 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: Threshold wavelength of the radiation (λ0)(\lambda _{0})= 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, threshold frequency (ν0\nu _{0} ) of the metal = 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 for 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 lambdaE=\frac{hc}{\ lambda })

Substituting these values in the expression for  lambda{\ lambda }:

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 nt=5n_{t}=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 = ∞ = Energy required for n1 = 1 to n = ∞, is 8.72×1020m8.72\times 10^{-20}\, m ΔE=EE5=[((2.18×1018J)(1)2()2)((2.18×1018J)(1)2(1)2)]=2.18×1018JΔE=2.18×1018J\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, a lower amount of energy is required in order to ionize electrons in the 5th orbital of a hydrogen atom when compared to do that in the ground state of the atom.

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?

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

Total no. of spectral lines emitted when an electron initially in the ‘nth‘ level drops down to the ground state can be calculated using the following expression:

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

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.

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

E5=(2.18×1018)(5)2=(2.18×1018)25=8.72×1020JE_{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, 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 possivle 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.

The ground-state electron energy is –2.18 × 10–11 ergs.

Ans. Energy (E) associated with the nth Bohr orbit of an atom is:

E5=(2.18×1018)Z2(n)2E_{5}=\frac{-(2.18\times 10^{-18})Z^{2}}{(n)^{2}}

Where, Z denotes the atom’s atomic number

Ground state energy =2.18×1011ergs=2.18×1011×107J=2.18×1018J=-2.18 \times 10^{–11}\, ergs\\ =-2.18 \times 10^{–11} \times 10^{–7}\, J \\ =-2.18 \times 10^{–18} \, J

The required energy for an electron shift from n = 1 to n = 5 is:

ΔE=E5E1=[((2.18×1018J)(1)2(5)2)(2.18×1018)]=(2.18×1018)[1125]=(2.18×1018)[2425]=2.0928×1018J\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

The wavelength of the emitted light = hcE \frac{hc}{E } E=(6.626×1034)(3×108)(2.0928×1018)=9.498×108mE=\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 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=0.545×1018J\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 ΔE=5.45×1019J\Delta E=5.45\times 10^{-19}\, J λ=hcΔE\lambda =\frac{hc}{\Delta E}

If λ is the longest wavelength that can cause this transition,

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=(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 thr 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 in the expression for λ:

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 associated with the electron which is 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. As per de Broglie’s equation,

λ=hmv\lambda =\frac{h}{mv }

Given, K.E of electron = 3.0×10253.0\times 10^{-25}J

Since K.E.= 12mv2\frac{1}{2}mv^{2} Velocity(v)=2K.Em=2(3.0×1025J)9.10939×1031kg=6.5866×104v=811.579ms1∴ 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 these values in the expression for λ:

λ=(6.626×1034)Js(9.10939×1031kg)(811.579ms1)=8.9625×107m\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)O2–

(d)F –

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

(a) 3s1

(b) 2p3 and

(c) 2p5?

(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

Electron 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 6

(d) F – ion

Electronic configuration of Fluorine = 1s22s22p5. The species has gained one electron (accounted for by the -1 charge). Electron 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 no. 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 no. electrons in the atom =2 + 2 + 3 = 7

the element’s atomic number is 7

(c) 2p 5

Complete electronic configuration: 1s 2 2s 2 2p 5 .

Total no. electrons in the atom = 2 + 2 + 5 = 9

the element’s atomic number is 9

(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.23. 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.24. 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.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)

In a neutral atom, no.protons = no.electrons. No. 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.26.Give the number of electrons in the species , H2+ and H2 and O2+

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

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

No. electrons O2 = 8 + 8 = 16. Number of electrons in O2+= 16 – 1 = 15

Q.27. (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.28. Usings, 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 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.29. 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 = 0, ml= 1, m=-12\frac{1}{2}

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

f) n = 3, l = 0, ml= 1, 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.

(f) Possible.

Q.30. 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 no. electrons = 2 (4)2 = 32

Electron configuration for an atom with 32 electrons: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 .

Hence, all the electrons are paired.

No. electrons (having n = 4 and m=-12\frac{1}{2}) = 16

(b)n = 3, l = 0 indicates the 3s orbital. Therefore, no. electrons with 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. Hydrogen atoms 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) 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.32. 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

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 that of n2 = 2.

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

Q.33. 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 process given by:

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

An electron is moved 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.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 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.

No. 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.35. 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

No. 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.36. 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)Radius of carbon atom= Diameter2=2.62=1.3×1010m=130×1012m=130pm\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

=1.6×102m 1.6 \times 10^{-2}m

Diameter of a zinc atom =1.6×1010m 1.6 \times 10^{-10}m No. zinc atoms in the arrangement

=1.6×102m2.6×1010m=0.6153×108m=6.153×107=\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 a 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 zinc atom= Diameter2=2.62=1.3×1010m=130×1012m=130pm\frac{Diameter}{2}\\ \\ =\frac{2.6}{2}\\ \\ =1.3\times 10^{-10}m\\ \\ =130\times 10^{-12}m=130pm

(b) )Length of entire arrangement = 1.6 cm

=1.6×102m 1.6 \times 10^{-2}m

Diameter of the zinc atom =1.6×1010m 1.6 \times 10^{-10}m

Therefore,  No. zinc atoms in the entire arrangement

=1.6×102m2.6×1010m=0.6153×108m=6.153×107=\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 2.5×1016C 2.5 \times 10^{-16}C  of static electric charge. Calculate the number of electrons present in it.

Ans.

Charge held by one electron = 1.6022×1019C 1.6022 \times 10^{-19}C 1.6022×1019C\Rightarrow 1.6022 \times 10^{-19}C charge is held by one electron.

Therefore, No. electrons that carry a charge of 2.5×1016C 2.5 \times 10^{-16}C 11.6022×1019C(2.5×1016C)=1.560×103C=1560C\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 is1.282×1018C -1.282 \times 10^{-18}C  , calculate the number of electrons present on it.

Ans.

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

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

Therefore, No. electrons present in the drop of oil

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.

The results obtained when a foil of heavy atoms will be different from the results obtained when relatively light atoms are used in the foil. The lighter the atom, the lower the magnitude of positive charge in its nucleus. Therefore, lighter atoms will not cause enough deflection of the positively charged α-particles.

 

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}.

Therefore, 3579Br_{35}^{79}\textrm{Br} is acceptable but 7935Br_{79}^{35}\textrm{Br} is not.

79Br _{}^{79}\textrm{Br} is an acceptable representation but 35Br _{}^{35}\textrm{Br} is not since the atomic numbers of elements are constant but mass numbers are not (due to the existence isotopes).

Q.42. An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.

Ans.

Let the No. protons in the element be x.

Therefore, No. neutrons in the element = x + 31.7% of x

= x + 0.317 x

= 1.317 x

Given, Mass number of the element = 81, which implies that (No. protons + No. neutrons) = 81

x+1.317x=812.317x=81x=812.317=34.95\Rightarrow x+1.317x=81\\ \\ 2.317x=81\\ \\ x=\frac{81}{2.317}\\ \\ =34.95\\ \\ x35x\simeq 35

Therefore, total no. protons = 35, which implies that atomic number = 35.

Therefore, the element is 3581Br_{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 no. electrons in the negatively charged ion be x.

Then, no. neutrons present = x + 11.1% of x = x + 0.111 x = 1.111 x

No. electrons present in the neutral atom = (x – 1)

(When an ion carries a negative charge, it carries an extra electron)

No. protons present in the neutral atom = x – 1

Given, mass number of the ion = 37

(x – 1) + 1.111x = 37

2.111x = 38

x = 18

Therefore, The symbol of the ion is 1737Cl_{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 total no. electrons present in A3+A^{3+} be x. Now, total no. neutrons in it = x + 30.4% of x = 1.304 x

Since the ion has a charge of +3 , \Rightarrow no. electrons in neutral atom = x + 3

Therefore, no. protons in neutral atom = x + 3

The Mass number of the ion is 56 (Given)

Therefore, (x+3)(1.304x)=56

2.304x=53

x=532.304x=\frac{53}{2.304}

X=23

Therefore, no. protons = x + 3 = 23 + 3 = 26

The ion is 2656Fe+_{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) X-rays.

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 a wavelength is as follows:

Cosmic rays < X-rays < radiation from microwave ovens < amber light < radiation of FM radio

 

Q.46. Nitrogen laser produces radiation at a wavelength of 337.1 nm. If the number of photons emitted is 5.6×10245.6 \times 10^{24}, calculate the power of this laser.

Ans.

Power of laser = Energy with which it emits photons

Power==E=Nhcλ=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=(5.6×1024)(6.626×1034Js)(3×108ms1)(337.1×109m)=0.3302×107J=3.33×106JE=\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 3.33×106J3.33\times 10^{6}J

 

Q.47. Neon gas is generally used in the signboards. If it emits strongly at 616 nm, calculate (a) the frequency of emission,

(b) distance travelled by this radiation in 30 s

(c) the energy of quantum and

(d) the number of quanta presents if it produces 2 J of energy.

Ans.

Wavelength of the emitted radiation = 616 nm =616×109m616\times 10^{-9}m(Given)

(a)Frequency of the emission (ν\nu)

ν=cλ\nu =\frac{c}{\lambda }

Where, c = speed of the radiation

λ = wavelength of the radiation

Substituting these values in the expression for (ν\nu):

ν=3×108m/s616×109m=4.87×108×109×103s1v=4.87×1014s1\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 the emission (ν\nu)= 4.87×1014s14.87\times 10^{14}s^{-1}

(b) Speed of the radiation,c =3×108ms13\times 10^{8}ms^{-1}

Distance travelled by the radiation in a timespan of 30 s

=(3×108ms1)(30s)=9×109m=(3\times 10^{8}ms^{-1})(30s)\\ \\ =9\times 10^{9}m

(c) Energy of one quantum (E) = hν

=(6.626×1034Js)(4.87×1014s1)=(6.626\times 10^{-34}Js)(4.87\times 10^{14}s^{-1})

Energy of one  quantum (E) =32.27×1020J32.27\times 10^{-20}J

Therefore, 32.27×1020J32.27\times 10^{-20}J of energy is present in 1 quantum.

No. quanta in 2 J of energy

2J32.27×1020J=6.19×1018=6.2×1018\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 3.15×1018J3.15\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=hcλE=\frac{hc}{\lambda }

Where,

λ denotes the wavelength of the radiation

h is Planck’s constant

c denotes the velocity of the radiation

Substituting these values in the expression for E:

E=(6.626×1034Js)(3×108ms1)(600×109)=3.313×1019JE=\frac{(6.626\times 10^{-34}Js)(3\times 10^{8}ms^{-1})}{(600\times 10^{-9})}=3.313\times 10^{-19}\, J

Energy held by one photon =3.313×1019J3.313\times 10^{-19}\, J

No. photons received with 3.15×1018J3.15\times 10^{-18}\, Jenergy

=3.15×1018J3.313×1019J=9.510=\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 nanosecond range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is 2.5×1015J2.5\times 10^{15}\, J, calculate the energy of the source.

Ans.

Frequency of radiation (ν\nu),

ν=12.0×109sν=5.0×108s1\nu =\frac{1}{2.0\times 10^{-9}s}\\ \\ \nu =5.0\times 10^{8}s^{-1}

Energy (E) of source = Nhν

Where,

N is the no. photons emitted

h is Planck’s constant

ν denotes the frequency of the radiation

Substituting these values in the expression for (E):

E=(2.5×1015)(6.626×1034Js)(5.0×108s1)E=8.282×1010JE=(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 8.282×1010J8.282\times 10^{-10}J.

 

Q.51. The work function for the 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 of 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.

Ans.

Given, the work function of caesium (W0) = 1.9 eV.

(a)From the W0=hcλ0W_{0}=\frac{hc}{\lambda _{0}}