1. Home
  2. |  
  3. CBSE Board |  

NCERT Solutions for CBSE Class 12 Physics ‒ Chapter 12: Atoms

Jun 6, 2017 18:20 IST

    Download NCERT Solutions for CBSE Class 12 Physics ‒ Chapter 12: Atoms

    NCERT Solutions for Class 12 Physics, Chapter 12: Atoms are available here. You can also download these solutions in PDF format. Some important topics of this chapter are Alpha-particle scattering and Rutherford’s nuclear model of atom, atomic spectra, Alpha-particle trajectory, Electron orbits, Spectral series, Bohr model of the hydrogen atom, line spectra of the hydrogen atom, De Broglie’s explanation of Bohr’s second postulate of quantisation. Most of the questions given in this chapter of NCERT textbook are based on these topics. These topics are also important for CBSE board exams and other competitive exams like NEET, JEE Mains etc.

    NCERT Solutions for Class 12 Physics ‒ Chapter 12: Atoms are given below:

    Question 12.1: Choose the correct alternative from the clues given at the end of the each statement:

    (a) The size of the atom in Thomson’s model is .......... the atomic size in Rutherford’s model. (much greater than/no different from/much less than)

    (b) In the ground state of .......... electrons are in stable equilibrium, while in .......... electrons always experience a net force. (Thomson’s model/ Rutherford’s model)

    (c) A classical atom based on .......... is doomed to collapse. (Thomson’s model/ Rutherford’s model)

    (d) An atom has a nearly continuous mass distribution in a ..........but has a highly non-uniform mass distribution in.......... (Thomson’s model/ Rutherford’s model)

    (e) The positively charged part of the atom possesses most of the mass in.......... (Rutherford’s model/both the models)

    Solution 12.1:

    (a) No different from

    (b) Thomson’s model; Rutherford’s model

    (c) Rutherford’s model

    (d) Thomson’s model; Rutherford’s model

    (e) Both the models

    NCERT Exemplar Questions & Solutions: CBSE Class 12 Physics – Chapter 12

    Question 12.2: Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect?

    Solution 12.2:

    There would be no large-angle scattering, because the alpha-particle has mass nearly four times than that of the target nuclei. No observable conclusion can be made in this solid hydrogen alpha-particle scattering experiment.

    Question 12.3: What is the shortest wavelength present in the Paschen series of spectral lines?

    Solution 12.3:

    Paschen series is given by

    (1/ λ) = R [(1/32) ‒ (1/n2)]

    For shortest wavelength, n = ∞

    Substituting and solving, λ = 820 nm.

    Question 12.4: A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level?

    Solution 12.4:

    Given:

    E = 2.3 eV

    We know

    v = E / h

    Substitution produces, v = 5.55 x 1014 Hz.

    Question 12.5: The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the electron in this state?

    Solution 12.5:

    Given:

    Total energy = ‒13.6 eV

    Kinetic energy = ‒ (Total Energy) = ‒ (‒13.6) = 13.6 eV

    Potential Energy = ‒ 2 (Kinetic energy) = ‒ 2 × 13.6 = ‒ 27.2 eV.

    Question 12.6: A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon.

    Solution 12.6:

    Given:

    n1 = 1

    n2 = 4

    Using the formula

    1/λ = R [(1/ni2) ‒ (1/nf2)]

    �� = 9.72 x 10‒8 m

    Frequency, v = c / ��= 3.1 x 1015 Hz.

     

    Question 12.7: (a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels.

    (b) Calculate the orbital period in each of these levels.

    Solution 12.7:

    (a) Bohr’s angular postulate momentum gives

    m vn rn = nh/2π  ...(1)

    Radius is given by,

    rn = (n2/m) × [h/2π]2 × [(4πϵo)/e2]  …(2)

    where symbols carry usual meanings.

    For n = 1

    r1 = 5.29 x 10-11 m

    Substituting in eq. (1)

    v1 = 2.19 x 106 m / s

    For n = 2

    Calculating radius using eq. (2)

    r2 = 2.12 x 10-10 m

    Putting in eq. (1)

    v2 = 1.09 x 106 m / s

    For n = 3

    Calculating radius using eq. (2)

    r3 = 4.78 x 10-10 m

    Putting in eq. (1) gives

    v3 = 7.27 x 105 m/s

    (b) Orbital period can be deduced by the simple relation of = distance / speed = (2 π rn)/vn …3

    For n = 1,

    Taking the corresponding values of r1 and v1 calculated above and substituting them in eq. (3)

    Orbital period = 1.52 x 10-16 s

    For n = 2

    In similar fashion

    Orbital period = 1.22 x 10-15 s

    For n = 3

    Orbital period = 4.11 x 10-15 s

    Question 12.8: The radius of the innermost electron orbit of a hydrogen atom is 5.3×10–11 m. What are the radii of the n = 2 and n =3 orbits?

    Solution 12.8: From the previous question,

    rn = (n2/m) × [h/2π]2 × [(4πϵo)/e2]

    Therefore

    For n = 2

    r2 = 2.12 x 10-10 m

    For n = 3

    r3 = 4.78 x 10-10 m

    Question 12.9: A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

    Solution 12.9: From the formula

    En = ‒13.6/n2 eV

    For n = 3 to n = 1

    E3 – E1 = 12.09 eV or (near to) 12.5 eV

    This energy difference includes Lyman and Balmer series.

    At room temperature, most of the hydrogen atoms are in ground state. So, n = 1, and the energy difference indicates the final n to be 2 or 3.

    Lyman series

    For n = 2

    (1/λ) = R [(1/12) ‒ (1/22)]

    λ = 122 nm

    For n = 3, ��= 103 nm

    Balmer series,

    For n = 3,

    1/λ = R [(1/22) ‒ (1/32)]

    ⇒ λ = 656 nm.

    Question 12.10: In accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution around the sun in an orbit of radius 1.5 × 1011 m with orbital speed 3 × 104 m/s. (Mass of earth = 6.0 × 1024 kg.)

    Solution 12.10:

    Given:

    r = 1.5 x 1011 m

    v = 30000 m / s

    m = 6 × 1024 kg

    Using the expression

    mvr = nh / 2 �� Substituting and solving

    n = 2.56 x 1074.

    NCERT Solutions for Class 12 Physics ‒ Chapter 12: Atoms

    Latest Videos

    Register to get FREE updates

      All Fields Mandatory
    • (Ex:9123456789)
    • Please Select Your Interest
    • Please specify

    • ajax-loader
    • A verifcation code has been sent to
      your mobile number

      Please enter the verification code below

    This website uses cookie or similar technologies, to enhance your browsing experience and provide personalised recommendations. By continuing to use our website, you agree to our Privacy Policy and Cookie Policy. OK
    X

    Register to view Complete PDF