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NEET-XII-Physics

43: Bohr's Model and Physics of the Atom

with Solutions - page 3

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  • Qstn #4
    Let An be the area enclosed by the nth orbit in a hydrogen atom. The graph of ln (An/A1) against ln(n)
    (a) will pass through the origin
    (b) will be a straight line with slope 4
    (c) will be a monotonically increasing nonlinear curve
    (d) will be a circle
    digAnsr:   a,b
    Ans : (a) will pass through the origin
    (b) will be a straight line with slope 4
    The radius of the nth orbit of a hydrogen atom is given by
    `` {r}_{\,\mathrm{\,n\,}}={n}^{2}{a}_{0}``
    Area of the nth orbit is given by
    `` {A}_{\,\mathrm{\,n\,}}=\,\mathrm{\,\pi \,}{r}_{\,\mathrm{\,n\,}}^{2}=\,\mathrm{\,\pi \,}{n}^{4}{a}_{0}^{2}``
    `` {A}_{1}=\,\mathrm{\,\pi \,}{a}_{0}^{2}``
    `` \Rightarrow \,\mathrm{\,ln\,}\left(\frac{{A}_{\,\mathrm{\,n\,}}}{{A}_{1}}\right)=\,\mathrm{\,ln\,}\left(\frac{\,\mathrm{\,\pi \,}{n}^{4}{a}_{0}^{2}}{\,\mathrm{\,\pi \,}{a}_{0}^{2}}\right)``
    `` \,\mathrm{\,ln\,}\left(\frac{{A}_{\,\mathrm{\,n\,}}}{{A}_{1}}\right)=4\,\mathrm{\,ln\,}n...\left(1\right)``
    From the above expression, the graph of ln (An/A1) against ln(n) will be a straight line passing through the origin and having slope 4.
    Page No 384:
  • Qstn #5
    Ionization energy of a hydrogen-like ion A is greater than that of another hydrogen-like ion B. Let r, u, E and L represent the radius of the orbit, speed of the electron, energy of the atom and orbital angular momentum of the electron respectively. In ground state
    (a) rA > rB
    (b) uA > uB
    (c) EA > EB
    (d) LA > LB
    digAnsr:   b,a
    Ans : (b) uA > uB
    The ionisation energy of a hydrogen like ion of atomic number Z is given by
    `` V=(13.6\,\mathrm{\,eV\,})\times {Z}^{2}``
    Thus, the atomic number of ion A is greater than that of B (ZA > ZB).
    The radius of the orbit is inversely proportional to the atomic number of the ion.
    ∴ rA > rB
    Thus,
    (a) is incorrect.
    The speed of electron is directly proportional to the atomic number.
    Therefore, the speed of the electron in the orbit of A will be more than that in B.
    Thus, uA > uB is correct.
    The total energy of the atom is given by
    `` E=-\frac{m{Z}^{2}{e}^{2}}{8{\in }_{0}{h}^{2}{n}^{2}}``
    As the energy is directly proportional to Z2, the energy of A will be less than that of B, i.e. EA < EB.
    The orbital angular momentum of the electron is independent of the atomic number.
    Therefore, the relation LA > LB is invalid.
    Page No 384:
  • Qstn #6
    When a photon stimulates the emission of another photon, the two photons have
    (a) same energy
    (b) same direction
    (c) same phase
    (d) same wavelength
    digAnsr:   a,b,c,d
    Ans : (a) same energy
    (b) same direction
    (c) same phase
    (d) same wavelength
    When a photon stimulates the emission of another photon, the two photons have same energy, direction, phase, and wavelength or we can say that the two photons are coherent.
    When an atom is present in its excited state then if a photon of energy equal to the energy gap between the excited state and any lower stable state is incident on this atom then the atom transits from upper state to the lower stable state by emitting a photon of energy equal to the energy gap between the two states. It is called stimulated emission. The emitted photon and incident photon have same energy and hence same wavelength. Also these two photons will be in phase and in the same direction. This process of producing monochromatic and unidirectional light is called lasing action.
    Page No 384:
  • #
    Section : iv
  • Qstn #1
    The Bohr radius is given by
    a0=ε0h2πme2. Verify that the RHS has dimensions of length.
    Ans : The dimensions of ε0 can be derived from the formula given below:
    `` a=\frac{{\epsilon }_{0}{h}^{2}}{\,\mathrm{\,\pi \,}m{e}^{2}}=\frac{{\,\mathrm{\,A\,}}^{2}{\,\mathrm{\,T\,}}^{2}{\left({\,\mathrm{\,ML\,}}^{2}{\,\mathrm{\,T\,}}^{-1}\right)}^{2}}{{\,\mathrm{\,L\,}}^{2}{\,\mathrm{\,ML\,}}^{-2}\,\mathrm{\,M\,}{\left(\,\mathrm{\,AT\,}\right)}^{2}}``
    `` =\frac{{\,\mathrm{\,M\,}}^{2}{\,\mathrm{\,L\,}}^{2}{\,\mathrm{\,T\,}}^{-2}}{{\,\mathrm{\,M\,}}^{2}{\,\mathrm{\,L\,}}^{3}{\,\mathrm{\,T\,}}^{-2}}=\,\mathrm{\,L\,}``
    Clearly, a0 has the dimensions of length.
    Page No 384:
  • Qstn #2
    Find the wavelength of the radiation emitted by hydrogen in the transitions
    Ans : From Balmer empirical formula, the wavelength `` \left(\lambda \right)`` of the radiation is given by
    `` \frac{1}{\lambda }=R\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)``
    Here, R = Rydberg constant = 1.097`` \times ``107 m`` -1``
    n1 = Quantum number of final state
    n2 = Quantum number of initial state
  • #2-a
    n = 3 to n = 2,
    Ans : For transition from n = 3 to n = 2:
    Here,
    n1 = 2
    n2 = 3
    `` \frac{1}{\,\mathrm{\,\lambda \,}}=1.09737\times {10}^{7}\times \left(\frac{1}{4}-\frac{1}{9}\right)``
    `` \Rightarrow \,\mathrm{\,\lambda \,}=\frac{36}{5\times 1.09737\times {10}^{7}}``
    `` =6.56\times {10}^{-7}=656\,\mathrm{\,nm\,}``
  • #2-b
    n = 5 to n = 4 and
    Ans : For transition from n = 5 to n = 4:
    Here,
    n1 = 4
    n2 = 5
    `` \frac{1}{\lambda }=1.09737\times {10}^{-7}\left(\frac{1}{16}-\frac{1}{25}\right)``
    `` \Rightarrow \lambda =\frac{400}{1.09737\times {10}^{7}\times 9}``
    `` =4050\,\mathrm{\,nm\,}``
  • #2-c
    n = 10 to n = 9.
    Ans : For transition from n = 10 to n = 9:
    Here,
    n1 = 9
    n2 = 10
    `` \frac{1}{\lambda }=1.09737\times {10}^{7}\left(\frac{1}{81}-\frac{1}{100}\right)``
    `` \lambda =\frac{81\times 100}{19\times 1.09737\times {10}^{7}}``
    `` =38849\,\mathrm{\,nm\,}``
    Page No 384:
  • Qstn #3
    Calculate the smallest wavelength of radiation that may be emitted by
    Ans : Given:
    For the smallest wavelength, energy should be maximum.
    Thus, for maximum energy, transition should be from infinity to the ground state.
    ∴ n1 = 1
    n2= `` \infty ``
  • #3-a
    hydrogen,
    Ans : Wavelength of the radiation emitted `` \left(\lambda \right)`` is given by
    `` \frac{\mathit{1}}{\lambda }=R{Z}^{2}\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)``
    For hydrogen,
    Atomic number, Z = 1
    R = Rydberg constant = 1.097×107 m`` -``1
    On substituting the respective values,
    `` \lambda =\frac{1}{1.097\times {10}^{7}}=\frac{1}{1.097}\times {10}^{-7}``
    `` =0.911\times {10}^{-7}``
    `` =91.1\times {10}^{-9}=91\,\mathrm{\,nm\,}``
  • #3-b
    He+ and
    Ans : For He+,
    Atomic number, Z = 2
    Wavelength of the radiation emitted by He+ (`` \lambda ``) is given by
    `` \frac{\mathit{1}}{\lambda }=R{Z}^{2}\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)``
    `` \therefore \frac{1}{\lambda }={\left(2\right)}^{2}(1.097\times {10}^{7})\left(\frac{1}{{\left(1\right)}^{2}}-\frac{1}{{\left(\infty \right)}^{2}}\right)``
    `` \Rightarrow \lambda =\frac{91\,\mathrm{\,nm\,}}{4}=23\,\mathrm{\,nm\,}``
  • #3-c
    Li++.
    Ans : For Li++,
    Atomic number, Z = 3
    Wavelength of the radiation emitted by Li++ (`` \lambda ``) is given by
    `` \frac{\mathit{1}}{\lambda }=R{Z}^{2}\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)``
    `` \therefore \frac{1}{\lambda }={\left(3\right)}^{2}\times (1.097\times {10}^{7})\left(\frac{1}{{1}^{2}}-\frac{1}{{\infty }^{2}}\right)``
    `` \Rightarrow \lambda =\frac{91\,\mathrm{\,nm\,}}{{Z}^{2}}=\frac{91}{9}=10\,\mathrm{\,nm\,}``
    Page No 384:
  • Qstn #4
    Evaluate Rydberg constant by putting the values of the fundamental constants in its expression.
    Ans : Expression of Rydberg constant (R) is given by
    `` R=\frac{m{e}^{4}}{8{h}^{2}c{\in }_{0}^{2}}``
    Mass of electron, me = 9.31`` \times ``1021 kg
    Charge, e = 1.6 × 10-19 C
    Planck's constant, h = 6.63 × 10-34 J-s,
    Speed of light, c = 3 × 108 m/s,
    Permittivity of vacuum, ∈0 = 8.85 × 10-12 C2N`` -1``m
    On substituting the values in the expression, we get
    `` R=\frac{\left(9.31\times {10}^{-31}\right)\times {\left(1.6\times {10}^{-19}\right)}^{4}}{8\times {\left(6.63\times {10}^{-34}\right)}^{2}\times \left(3\times {10}^{8}\right)\times {\left(8.85\times {10}^{-12}\right)}^{2}}``
    `` \Rightarrow R=1.097\times {10}^{7}{\,\mathrm{\,m\,}}^{-1}``
    Page No 384:
  • Qstn #5
    Find the binding energy of a hydrogen atom in the state n = 2.
    Ans : The binding energy (E) of hydrogen atom is given by
    `` E=\frac{-13.6}{{n}^{2}}\,\mathrm{\,eV\,}``
    For state n = 2,
    `` E=-\frac{13.6}{{\left(2\right)}^{2}}``
    `` \Rightarrow E=-3.4\,\mathrm{\,eV\,}``
    Thus, binding energy of hydrogen at n = 2 is `` -3.4\,\mathrm{\,eV\,}``.
    Page No 384:
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