43: Bohr's Model And Physics Of The Atom : Neet-xii-physics

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NEET-XII-Physics

43: Bohr's Model and Physics of the Atom

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Qstn #6
Find the radius and energy of a He⁺ ion in the states
Ans : For He⁺ ion,
Atomic number, Z = 2
For hydrogen like ions, radius `` \left(r\right)`` of the nth state is given by
`` r=\frac{0.53{n}^{2}}{Z}``Å
Here,
Z = Atomic number of ions
n = Quantum number of the state
Energy `` \left(E\right)`` of the nth state is given by
E_n = `` -\frac{13.6{Z}^{2}}{{n}^{2}}``

#6-a
n = 1,
Ans : For n = 1,
Radius, `` r=\frac{0.53\times {\left(1\right)}^{2}}{2}``
`` =0.265{\,\mathrm{\,A\,}}^{0}`` Å
Energy, E_n = `` \frac{-13.6\times 4}{1}``
= `` -`` 54.4 eV

#6-b
n = 4 and
Ans : For n = 4,
Radius, `` r=\frac{0.53\times 16}{2}=4.24\,\mathrm{\,\AA \,}``
Energy, `` E=\frac{-13.6\times 4}{16}=-3.4e\,\mathrm{\,V\,}``

#6-c
n = 10.
Ans : For n = 10,
Radius, `` r=\frac{0.53\times 100}{2}``
= 26.5 Å
Energy, `` E=\frac{-13.6\times 4}{100}=-0.544\,\mathrm{\,eV\,}``
Page No 384:

Qstn #7
A hydrogen atom emits ultraviolet radiation of wavelength 102.5 nm. What are the quantum numbers of the states involved in the transition?
Ans : Wavelength of ultraviolet radiation, `` \lambda `` = 102.5 nm = 102,5 `` \times ``10^{`` -9``} m
Rydberg's constant, R = 1.097`` \times `` 10⁷ m^{`` -1``}
Since the emitted light lies in ultraviolet range, the lines will lie in lyman series.
Lyman series is obtained when an electron jumps to the ground state (n₁= 1) from any excited state (n₂).
Wavelength of light `` \left(\lambda \right)`` is given by
`` \frac{1}{\lambda }=R\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)``
`` \,\mathrm{\,Here\,},R=\,\mathrm{\,Rydberg\,}\,\mathrm{\,constant\,}``
`` \Rightarrow \frac{1}{102.5\times {10}^{-9}}=1.097\times {10}^{7}\left(\frac{1}{{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)``
`` \Rightarrow \frac{{10}^{9}}{102.5}=1.097\times {10}^{7}\left(1-\frac{1}{{n}_{2}^{2}}\right)``
`` \Rightarrow \frac{{10}^{2}}{102.5}=1.097\left(1-\frac{1}{{n}_{2}^{2}}\right)``
`` \Rightarrow 1-\frac{1}{{n}_{2}^{2}}=\frac{100}{102.5\times 1.097}``
`` \Rightarrow \frac{1}{{n}_{2}^{2}}=1-\frac{100}{102.5\times 1.097}``
`` \Rightarrow {n}_{2}=\sqrt{9.041}=3``
The transition will be from 1 to 3.
Page No 384:

#8-a
Find the first excitation potential of He⁺ ion.
Ans : PE of hydrogen like atom in the nth state, V = `` =\frac{-13.6{Z}^{2}}{{n}^{2}}\,\mathrm{\,eV\,}``
`` ``
Here, Z is the atomic number of that atom.
For the first excitation, the atom has to be excited from n = 1 to n = 2 state.
So, its excitation potential will be equal to the difference in the potential of the atom in n = 1 and in n = 2 states.
First excitation potential of He⁺
`` -13.6{Z}^{2}(1-\frac{1}{{2}^{2}})eV``
`` =-10.2{Z}^{2}eV``
⇒10.2 × Z²
= 10.2 × 4
= 40.8 V

#8-b
Find the ionization potential of Li⁺⁺ ion.
Ans : Ionization Potential Li⁺⁺ = 13.6 V × Z²
= 13.6 × 9
= 122.4 V
Page No 384:

Qstn #9
A group of hydrogen atoms are prepared in n = 4 states. List the wavelength that are emitted as the atoms make transitions and return to n = 2 states.
Ans : There will be three wavelengths.
(i) For the transition from (n = 4) to (n = 3) state
(ii) For the transition from (n = 3) to (n = 2) state
(iii) For the transition from (n = 4) to (n = 2) state
Let (`` \lambda ``₁) be the wavelength when the atom makes transition from (n = 4) state to (n = 2) state.
Here,
n₁= 2
n₂= 4
Now, the wavelength (`` \lambda ``₁) will be
`` \frac{1}{{\lambda }_{1}}=R\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)``
`` ``
`` ``
`` R=1.097\times {10}^{7}{\,\mathrm{\,m\,}}^{-1}``
`` ``
`` \frac{1}{{\lambda }_{1}}=1.097\times {10}^{7}\times \left(\frac{1}{4}-\frac{1}{16}\right)``
`` \Rightarrow \frac{1}{{\lambda }_{1}}=1.097\times {10}^{7}\left(\frac{4-1}{16}\right)``
`` \Rightarrow \frac{1}{{\lambda }_{1}}=\frac{1.097\times {10}^{7}\times 3}{16}``
`` \Rightarrow {\lambda }_{1}=\frac{16\times {10}^{-7}}{3\times 1.097}``
`` =4.8617\times {10}^{-7}``
`` =486.1\times {10}^{-9}``
`` =487\,\mathrm{\,nm\,}``
When an atom makes transition from (n = 4) to (n = 3), the wavelength (`` \lambda ``2) is given by
`` \,\mathrm{\,Here\,}\,\mathrm{\,again\,},``
`` {n}_{1}=3``
`` {n}_{2}=4``
`` \frac{1}{{\lambda }_{2}}=1.097\times {10}^{7}\left(\frac{1}{9}-\frac{1}{16}\right)``
`` \Rightarrow \frac{1}{{\lambda }_{2}}=1.097\times {10}^{7}\left(\frac{16-9}{144}\right)``
`` \Rightarrow \frac{1}{{\lambda }_{2}}=\frac{1.097\times {10}^{7}\times 7}{144}``
`` \Rightarrow {\lambda }_{2}=\frac{144}{1.097\times {10}^{7}\times 7}``
`` =1875\,\mathrm{\,nm\,}``
Similarly, wavelength (`` \lambda ``₃) for the transition from (n = 3) to (n = 2) is given by
When the transition is n₁ = 2 to n₂ = 3:
`` \frac{1}{{\lambda }_{3}}=1.097\times {10}^{7}\left(\frac{1}{4}-\frac{1}{9}\right)``
`` \Rightarrow \frac{1}{{\lambda }_{3}}=1.097\times {10}^{7}\left(\frac{9-4}{36}\right)``
`` \Rightarrow \frac{1}{{\lambda }_{3}}=\frac{1.097\times {10}^{7}\times 5}{36}``
`` \Rightarrow {\lambda }_{3}=\frac{36\times {10}^{-7}}{1.097\times 5}=656\,\mathrm{\,nm\,}``
Page No 384:

Qstn #10
A positive ion having just one electron ejects it if a photon of wavelength 228 Å or less is absorbed by it. Identify the ion.
Ans : Given:
Wavelength of photon, λ = 228 Å
Energy `` \left(E\right)`` is given by
`` E=\frac{h\,\mathrm{\,c\,}}{\lambda }``
`` ``
Here, c = Speed of light
h = Planck's constant
`` \therefore E=\frac{\left(6.63\times {10}^{-34}\right)\times \left(3\times {10}^{8}\right)}{\left(228\times {10}^{-10}\right)}``
`` =0.0872\times {10}^{-16}\,\mathrm{\,J\,}``
As the transition takes place from n = 1 to n = 2, the excitation energy (E₁) will be
`` {E}_{1}=Rh\,\mathrm{\,c\,}{Z}^{2}\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)``
`` {E}_{1}=\left(13.6\,\mathrm{\,eV\,}\right)\times {Z}^{2}\times \left(\frac{1}{{1}^{2}}-\frac{1}{{2}^{2}}\right)``
`` \Rightarrow {E}_{1}=\left(13.6\,\mathrm{\,eV\,}\right)\times {Z}^{2}\times \frac{3}{4}``
This excitation energy should be equal to the energy of the photon.
`` \therefore 13.6\times \frac{3}{4}\times {\,\mathrm{\,Z\,}}^{2}=0.0872\times {10}^{-16}``
`` {Z}^{2}=\frac{0.0872\times {10}^{-16}\times 4}{13.6\times 3\times 1.6\times {10}^{-19}}=5.34``
`` Z=\sqrt{5.34}=2.3``
The ion may be helium.
Page No 384:

Qstn #11
Find the maximum Coulomb force that can act on the electron due to the nucleus in a hydrogen atom.
Ans : Charge on the electron, q₁ = 1.6`` \times ``10 ^{`` -19``}C
Charge on the nucleus, q₂ = 1.6`` \times ``10 ^{`` -19``}C
Let r be the distance between the nucleus and the electron.
Coulomb force `` \left(F\right)`` is given by
`` F=\frac{{q}_{1}{q}_{2}}{4\,\mathrm{\,\pi \,}{\in }_{0}{r}^{2}}`` .....(1)
Here, q₁ = q₂ = q = 1.6`` \times ``10 ^{`` -19``}C
Smallest distance between the nucleus and the first orbit, r = 0.53`` \times ``10^{`` -``10} m
`` K=\frac{1}{4{\,\mathrm{\,\pi \epsilon \,}}_{0}}`` = 9`` \times ``10⁹ Nm²C^{`` -``2}
Substituting the respective values in (1), we get
`` F=\frac{\left(9\times {10}^{9}\right)\times \left(1.6\times {10}^{-19}\right)\times \left(1.6\times {10}^{-19}\right)}{{\left(0.53\times {10}^{-10}\right)}^{2}}``
`` =\frac{1.6\times 1.6\times 9\times {10}^{-9}}{{\left(0.53\right)}^{2}}=82.02\times {10}^{-9}``
`` =8.2\times {10}^{-8}\,\mathrm{\,N\,}``
Page No 384:

Qstn #12
A hydrogen atom in a state having a binding energy of 0.85 eV makes transition to a state with excitation energy 10.2 e.V

#12-a
Identify the quantum numbers n of the upper and the lower energy states involved in the transition.
Ans : The binding energy of hydrogen is given by
`` E=\frac{13.6}{{n}^{2}}\,\mathrm{\,eV\,}``
For binding energy of 0.85 eV,
`` {{n}_{2}}^{2}=\frac{13.6}{0.85}=16``
`` {n}_{2}=4``
For binding energy of 10.2 eV,
`` {{n}_{1}}^{2}=\frac{13.6}{10.2}``
`` {n}_{1}=1.15``
`` \Rightarrow {n}_{1}=2``
`` ``
The quantum number of the upper and the lower energy state are 4 and 2, respectively.

#12-b
Find the wavelength of the emitted radiation.
Ans : Wavelength of the emitted radiation `` \left(\lambda \right)`` is given by
`` \frac{1}{\lambda }=R\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)``
Here,
R = Rydberg constant
n₁ and n₂ are quantum numbers.
`` \therefore \frac{1}{\lambda }=1.097\times {10}^{7}\left(\frac{1}{4}-\frac{1}{16}\right)``
`` \Rightarrow \,\mathrm{\,\lambda \,}=\frac{16}{1.097\times 3\times {10}^{7}}``
`` =4.8617\times {10}^{-7}``
`` =487\,\mathrm{\,nm\,}``
Page No 384:

Qstn #13
Whenever a photon is emitted by hydrogen in Balmer series, it is followed by another photon in Lyman series. What wavelength does this latter photon correspond to?
Ans : As the second photon emitted lies in the Lyman series, the transition will be from the states having quantum numbers n = 2 to n = 1.
Wavelength of radiation `` \left(\lambda \right)`` is given by
`` \frac{1}{\lambda }=R\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)``
`` ``
Here, R is the Rydberg constant, having the value of 1.097×10⁷ m^-1.
`` \frac{1}{\lambda }=1.097\times {10}^{7}\left[\frac{1}{{\left(1\right)}^{2}}-\frac{1}{{\left(2\right)}^{2}}\right]``
`` \Rightarrow \frac{1}{\lambda }=1.097\times {10}^{7}\left[1-\frac{1}{4}\right]``
`` \Rightarrow \frac{1}{\lambda }=1.097\times \frac{3}{4}\times {10}^{7}``
`` \Rightarrow \lambda =\frac{4}{1.097\times 3\times {10}^{7}}``
`` =1.215\times {10}^{-7}``
`` =121.5\times {10}^{-9}=122\,\mathrm{\,nm\,}``
Page No 384: