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

Home Learn Fast select Course

Change Unit Change Subject Last Menu

NEET-XII-Physics

43: Bohr's Model and Physics of the Atom

with Solutions - page 5

Note: Please signup/signin free to get personalized experience.

10 minutes can boost your percentage by 10%

Note: Please signup/signin free to get personalized experience.

Qstn #14
A hydrogen atom in state n = 6 makes two successive transitions and reaches the ground state. In the first transition a photon of 1.13 eV is emitted. (a) Find the energy of the photon emitted in the second transition (b) What is the value of n in the intermediate state?
digAnsr: b
Ans : Energy (E) of the nth state of hydrogen atom is given by
E = `` \frac{13.6}{{n}^{2}}`` eV
For n = 6,
`` \therefore E=\frac{-13.6}{36}=-0.377777777`` eV
Energy of hydrogen atom in the ground state = -13.6 eV
Energy emitted in the second transition = Energy of ground state `` -`` (Energy of hydrogen atom in the 6th state + Energy of photon)
`` =13.6-\left(0.3777\overline{)7}+1.13\right)``
`` =12.09=12.1\,\mathrm{\,eV\,}``
(b)
Energy in the intermediate state = (Energy of photon emitted in the first transition) + (Energy of n = 6 state)
= 1.13 eV + 0.377 eV
= 1.507 eV
Energy of the nth state can expressed as
`` \frac{13.6}{{n}^{2}}=1.507``
`` ``
`` \therefore n=\sqrt{\frac{13.6}{1.507}}=\sqrt{9.024}\approx 3``
Page No 384:

Qstn #15
What is the energy of a hydrogen atom in the first excited state if the potential energy is taken to be zero in the ground state?
Ans : In ground state, the potential energy of a hydrogen atom is zero.
An electron is bound to the nucleus with an energy of 13.6 eV.
Therefore, we have to give 13.6 eV energy to move the electron from the nucleus.
Let us calculate the excitation energy required to take an atom from the ground state (n = 1) to the first excited state (n = 2).
`` E=13.6\times \left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)\,\mathrm{\,eV\,}``
Therefore, the excitation energy is given by
`` E=13.6\times \left(\frac{1}{{1}^{2}}-\frac{1}{{2}^{2}}\right)\,\mathrm{\,eV\,}``
`` E=13.6\times \frac{3}{4}\,\mathrm{\,eV\,}=10.2\,\mathrm{\,eV\,}``
Energy of 10.2 eV is needed to take an atom from the ground state to the first excited state.
∴ Total energy of an atom in the first excitation state = 13.6 eV + 10.2 eV = 23.8 eV
Page No 384:

Qstn #16
A hot gas emits radiation of wavelengths 46.0 nm, 82.8 nm and 103.5 nm only. Assume that the atoms have only two excited states and the difference between consecutive energy levels decreases as energy is increased. Taking the energy of the highest energy state to be zero, find the energies of the ground state and the first excited state.
Ans : Given:
Energy (E) of the ground state will be the energy acquired in the transition of the 2 excitation state to ground state.
`` {E}_{1}=\frac{hc}{{\,\mathrm{\,\lambda \,}}_{1}}``
`` ``
Here,
h = Planck's constant
c = Speed of light
`` {\lambda }_{1}`` = Wavelength of the radiation emitted when atoms come from the highest excited state to ground state
`` \therefore {E}_{1}=\frac{(6.63\times {10}^{-34})\times (3\times {10}^{8})}{(46\times {10}^{-9})}\,\mathrm{\,J\,}``
`` {E}_{1}=\frac{(6.63\times {10}^{-34})\times (3\times {10}^{8})}{(46\times {10}^{-9})\times (1.6\times {10}^{-19})}\,\mathrm{\,eV\,}``
`` =\frac{1242}{46}=27\,\mathrm{\,eV\,}``
Energy in the first excitation state `` \left({E}_{2}\right)`` will be the energy acquired in the transition of the highest energy state to the 2nd excitation state.
`` {E}_{2}=\frac{hc}{{\,\mathrm{\,\lambda \,}}_{n}}``
`` ``
Here,
`` {\lambda }_{n}`` = Wavelength of the radiation emitted when an atom comes from the highest energy state to the 2nd excitation state.
`` {E}_{2}=\frac{hc}{{\,\mathrm{\,\lambda \,}}_{n}}``
`` {E}_{2}=\frac{(6.63\times {10}^{-34})\times (3\times {10}^{8})}{(103.5\times {10}^{-9})}\,\mathrm{\,J\,}``
`` {E}_{2}=\frac{(6.63\times {10}^{-34})\times (3\times {10}^{8})}{(103.5\times {10}^{-9})\times (1.6\times {10}^{-19})}\,\mathrm{\,eV\,}``
`` =12\,\mathrm{\,eV\,}``
Page No 384:

Qstn #17
A gas of hydrogen-like ions is prepared in a particular excited state A. It emits photons having wavelength equal to the wavelength of the first line of the Lyman series together with photons of five other wavelengths. Identify the gas and find the principal quantum number of the state A.
Ans : (a) If the atom is excited to the principal quantum (n), then the number of transitions is given by
`` \frac{n\left(n-1\right)}{2}``
`` ``
It is given that a total of 6 photons are emitted. Therefore, total number of transitions is 6.
∴ `` \frac{n\left(n-1\right)}{2}``
`` `` = 6
⇒ n = 4
Thus, the principal quantum number is 4 and the gas is in the 4th excited state.
Page No 384:

Qstn #18
Find the maximum angular speed of the electron of a hydrogen atom in a stationary orbit.
Ans : Let the mass of the electron be m.
Let the radius of the hydrogen's first stationary orbit be r.
Let the linear speed and the angular speed of the electron be v and ω, respectively.
According to the Bohr's theory, angular momentum (L) of the electron is an integral multiple of h/2`` \,\mathrm{\,\pi \,}``, where h is the Planck's constant.
`` \Rightarrow mvr=\frac{nh}{2\,\mathrm{\,\pi \,}}`` (Here, n is an integer.)
`` v=r\omega ``
`` \Rightarrow m{r}^{2}\,\mathrm{\,\omega \,}=\frac{nh}{2\,\mathrm{\,\pi \,}}``
`` \Rightarrow \omega =\frac{nh}{2\,\mathrm{\,\pi \,}\times m\times {r}^{2}}``
`` ``
`` \therefore \omega =\frac{1\times \left(6.63\times {10}^{-34}\right)}{2\times \left(3.14\right)\times \left(9.1093\times {10}^{-31}\right)\times {\left(0.53\times {10}^{-10}\right)}^{2}}``
`` =0.413\times {10}^{17}\,\mathrm{\,rad\,}/\,\mathrm{\,s\,}=4.13\times {10}^{16}\,\mathrm{\,rad\,}/\,\mathrm{\,s\,}``
Page No 384:

Qstn #19
A spectroscopic instrument can resolve two nearby wavelengths λ and λ + Δλ if λ/Δλ is smaller than 8000. This is used to study the spectral lines of the Balmer series of hydrogen. Approximately how many lines will be resolved by the instrument?
Ans : The range of wavelength falling in Balmer series is between 656.3 nm and 365 nm.
It is given that the resolution of the instrument is λ/ (λ + ∆λ) < 8000.
Number of wavelengths in this range will be calculated in the following way:
`` \frac{656.3-365}{8000}=36``
Two lines will be extra for the first and last wavelength.
∴ Total number of lines = 36 + 2 = 38
Page No 384:

Qstn #20
Suppose, in certain conditions only those transitions are allowed to hydrogen atoms in which the principal quantum number n changes by 2.
Ans : Given:
Possible transitions:
From n₁ = 1 to n₂ = 3
n₁ = 2 to n₂ = 4

#20-a
Find the smallest wavelength emitted by hydrogen.
Ans : Here, n₁ = 1 and n₂ = 3
Energy, `` E=13.6\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)``
`` E=13.6\left(\frac{1}{1}-\frac{1}{9}\right)``
`` =13.6\times \frac{8}{9}.....\left(1\right)``
`` \,\mathrm{\,Energy\,}\left(E\right)\,\mathrm{\,is\,}\,\mathrm{\,also\,}\,\mathrm{\,given\,}\,\mathrm{\,by\,}``
`` E=\frac{hc}{\,\mathrm{\,\lambda \,}}``
`` ``
Here, h = Planck constant
c = Speed of the light
`` \lambda `` = Wavelength of the radiation
`` ``
`` \therefore E=\frac{6.63\times {10}^{-34}\times 3\times {10}^{8}}{\lambda }.....\left(2\right)``
`` \,\mathrm{\,Equating\,}\,\mathrm{\,equation\,}s\left(1\right)\,\mathrm{\,and\,}\left(2\right),\,\mathrm{\,we\,}\,\mathrm{\,have\,}``
`` \Rightarrow \,\mathrm{\,\lambda \,}=\frac{6.63\times {10}^{-34}\times 3\times {10}^{8}\times 9}{13.6\times 8}``
`` =0.027\times {10}^{-7}=103\,\mathrm{\,nm\,}``

#20-b
List the wavelength emitted by hydrogen in the visible range (380 nm to 780 nm).
Ans : Visible radiation comes in Balmer series.
As 'n' changes by 2, we consider n = 2 to n = 4.
Energy, `` {E}_{1}=13.6\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)``
`` =13.6\times \left(\frac{1}{4}-\frac{1}{16}\right)``
`` =2.55\,\mathrm{\,eV\,}``
If `` {\lambda }_{1}`` is the wavelength of the radiation, when transition takes place between quantum number n = 2 to n = 4, then
255 = `` \frac{1242}{{\lambda }_{1}}``
or λ₁ = 487 nm
Page No 384:

Qstn #21
According to Maxwell’s theory of electrodynamics, an electron going in a circle should emit radiation of frequency equal to its frequency of revolution. What should be the wavelength of the radiation emitted by a hydrogen atom in ground state if this rule is followed?
Ans : Let v₀ be the velocity of the electron moving in the ground state and `` {r}_{0}`` be the radius of the ground state.
Frequency of the revolution of electron in the circle is given by
`` f=\frac{{v}_{0}}{2\,\mathrm{\,\pi \,}{r}_{0}}``
Frequency of the radiation emitted = Frequency of the revolution of electron
∴ Frequency of the radiation emitted = `` \frac{{v}_{0}}{2\,\mathrm{\,\pi \,}{r}_{0}}``
Also, c = f`` \lambda ``
Here, c = Speed of light
`` \lambda `` = Wavelength of the radiation emitted
`` \Rightarrow `` `` \lambda =\frac{c}{f}``
`` \therefore \lambda =\frac{2\,\mathrm{\,\pi \,}{r}_{0}c}{{v}_{0}}``
`` =\frac{2\times \left(3.14\right)\times \left(53\times {10}^{-12}\right)\times (3\times {10}^{8})}{\left(2.187\times {10}^{6}\right)}``
`` =45.686\times {10}^{-12}\,\mathrm{\,m\,}=45.7\,\mathrm{\,nm\,}``
Page No 385:

Qstn #22
The average kinetic energy of molecules in a gas at temperature T is 1.5 kT. Find the temperature at which the average kinetic energy of the molecules of hydrogen equals the binding energy of its atoms. Will hydrogen remain in molecular from at this temperature? Take k = 8.62 × 10^-5 eV K^-1.
Ans : Average kinetic energy `` \left(K\right)`` of the molecules in a gas at temperature `` \left(T\right)`` is given by
K = `` \frac{3}{2}kT``
Here,
k = 8.62 × 10^-5 eVK^-1
T = Temperature of gas
The binding energy of hydrogen atom is 13.6 eV.
According to the question,
Average kinetic energy of hydrogen molecules = Binding energy of hydrogen atom
`` \therefore ``1.5 kT = 13.6
`` \Rightarrow ``1.5 × 8.62 × 10^-5 × T = 13.6
`` \Rightarrow T=\frac{13.6}{1.5\times 8.62\times {10}^{-5}}``
`` =1.05\times {10}^{5}\,\mathrm{\,K\,}``
No, it is impossible for hydrogen to remain in molecular state at such a high temperature.
Page No 385:

Qstn #23
Find the temperature at which the average thermal kinetic energy is equal to the energy needed to take a hydrogen atom from its ground state to n = 3 state. Hydrogen can now emit red light of wavelength 653.1 nm. Because of Maxwellian distribution of speeds, a hydrogen sample emits red light at temperatures much lower than that obtained from this problem. Assume that hydrogen molecules dissociate into atoms.
Ans : Given:
Wavelength of red light, `` \lambda `` = 653.1 nm = 653.1`` \times ``10^{`` -9``} m
Kinetic energy of H₂ molecules `` \left(K\right)`` is given by
`` K=\frac{3}{2}kT`` ...(1)
Here, k = 8.62 × 10^-5 eV/K
T = Temperature of H₂ molecules
Energy released `` \left(E\right)`` when atom goes from ground state to n = 3 is given by
`` E=13.6\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)``
For ground state, n₁ = 1
Also, n₂ = 3
`` \therefore E=13.6\left(\frac{1}{1}-\frac{1}{9}\right)``
`` =13.6\left(\frac{8}{9}\right)...\left(2\right)``
`` ``
`` ``
Kinetic energy of H₂ molecules = Energy released when hydrogen atom goes from ground state to n = 3 state
`` \therefore \frac{3}{2}\times 8.62\times {10}^{-5}\times T=\frac{13.6\times 8}{9}``
`` \Rightarrow T=\frac{13.6\times 8\times 2}{9\times 3\times 8.62\times {10}^{-5}}``
`` =9.4\times {10}^{4}\,\mathrm{\,K\,}``
Page No 385:

Qstn #24
Average lifetime of a hydrogen atom excited to n = 2 state is 10^-8 s. Find the number of revolutions made by the electron on the average before it jumps to the ground state.
Ans : Frequency of electron (d) is given by
`` f=\frac{m{e}^{4}}{4{{\in }_{0}}^{2}{n}^{3}{h}^{3}}``
`` ``
Time period is given by
`` T=\frac{1}{f}``
`` T=\frac{4{\in }_{0}^{2}{n}^{3}{h}^{3}}{m{e}^{4}}``
Here,
h = Planck's constant
m = Mass of the electron
e = Charge on the electron
`` {\epsilon }_{0}``= Permittivity of free space
`` \therefore T=\frac{4\times \left(8.85\times {10}^{-12}\right)\times {\left(2\right)}^{3}\times {\left(6.63\times {10}^{-34}\right)}^{3}}{\left(9.10\times {10}^{-31}\right)\times {\left(1.6\times {10}^{-16}\right)}^{4}}``
`` =12247.735\times {10}^{-19}\,\mathrm{\,s\,}``
Average life time of hydrogen, t = 10^{`` -``8} s
Number of revolutions is given by

`` N=\frac{t}{T}``
`` \Rightarrow N=\frac{{10}^{-8}}{12247.735\times {10}^{-19}}``
N = 8.2 `` \times `` 10⁵revolution
Page No 385:

Qstn #25
Calculate the magnetic dipole moment corresponding to the motion of the electron in the ground state of a hydrogen atom.
Ans : Mass of the electron, m = 9.1×10^{`` -``31}kg
Radius of the ground state, r = 0.53×10^{`` -``10}m
Let f be the frequency of revolution of the electron moving in ground state and A be the area of orbit.
Dipole moment of the electron (μ) is given by
μ = niA = qfA
`` =e\times \frac{m{e}^{4}}{4{\in }_{0}^{2}{h}^{3}{n}^{3}}\times \left(\,\mathrm{\,\pi \,}{r}^{2}{n}^{2}\right)``
`` =\frac{m{e}^{5}\times \left(\,\mathrm{\,\pi \,}{r}^{2}{n}^{2}\right)}{4{\in }_{0}^{2}{h}^{3}{n}^{3}}``
`` ``
Here,
h = Planck's constant
e = Charge on the electron
`` {\epsilon }_{0}`` = Permittivity of free space
n = Principal quantum number
`` \therefore \mu =\frac{\left(9.1\times {10}^{-31}\right){\left(1.6\times {10}^{-19}\right)}^{5}\times \,\mathrm{\,\pi \,}\times {\left(0.53\times {10}^{-10}\right)}^{2}}{4\times {\left(8.85\times {10}^{-12}\right)}^{2}\times {\left(6.64\times {10}^{-34}\right)}^{3}\times {\left(1\right)}^{3}}``
`` =0.000917\times {10}^{-20}``
`` =9.176\times {10}^{-24}\,\mathrm{\,A\,}-{\,\mathrm{\,m\,}}^{2}``
Page No 385:

Qstn #26
Show that the ratio of the magnetic dipole moment to the angular momentum (l = mvr) is a universal constant for hydrogen-like atoms and ions. Find its value.
Ans : Mass of the electron, m = 9.1×10^{`` -``31}kg
Radius of the ground state, r = 0.53×10^{`` -``10}m
Let f be the frequency of revolution of the electron moving in the ground state and A be the area of orbit.
Dipole moment of the hydrogen like elements (μ) is given by
μ = niA = qfA
`` =e\times \frac{m{e}^{4}}{4{\in }_{0}^{2}{h}^{3}{n}^{3}}\times \left(\,\mathrm{\,\pi \,}{r}_{0}^{2}{n}^{2}\right)``
`` =\frac{m{e}^{5}\times \left(\,\mathrm{\,\pi \,}{{r}_{0}}^{2}{n}^{2}\right)}{4{\in }_{0}^{2}{h}^{3}{n}^{3}}``
`` ``
Here,
h = Planck's constant
e = Charge on the electron
`` {\epsilon }_{0}`` = Permittivity of free space
n = Principal quantum number
Angular momentum of the electron in the hydrogen like atoms and ions (L) is given by
`` L=mvr=\frac{nh}{2\,\mathrm{\,\pi \,}}``
Ratio of the dipole moment and the angular momentum is given by
`` \frac{\mu }{L}=\frac{{e}^{5}\times m\times \,\mathrm{\,\pi \,}{r}^{2}{n}^{2}}{4{\in }_{0}{h}^{3}{n}^{3}}\times \frac{2\,\mathrm{\,\pi \,}}{nh}``
`` \frac{\mu }{L}=\frac{{\left(1.6\times {10}^{-19}\right)}^{5}\times \left(9.10\times {10}^{-31}\right)\times {\left(3.14\right)}^{2}\times {\left(0.53\times {10}^{-10}\right)}^{2}}{2\times {\left(8.85\times {10}^{-12}\right)}^{2}\times {\left(6.63\times {10}^{-34}\right)}^{3}\times {1}^{2}}``
`` \frac{\mu }{L}=3.73\times {10}^{10}\,\mathrm{\,C\,}/\,\mathrm{\,kg\,}``
Ratio of the magnetic dipole moment and the angular momentum do not depends on the atomic number 'Z'.
Hence, it is a universal constant.
Page No 385: