Atoms - Class 12 Engineering Physics - Extra Questions

Find the value of $$T \times 10^{17}$$ where $$T$$ is the orbital period of the electron?

$$T=\cfrac { 2\pi r }{ v } =\cfrac { 2\pi \times 5\times { 10 }^{ -11 } }{ \pi \times { 10 }^{ 6 } } =10\times { 10 }^{ -17 }sec$$
So, $$T=10$$


(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.

(a) $$v$$ is the orbital speed of the electron in the hydrogen atom in level $$n$$.
$$v=\dfrac{e^2}{n_14\pi\epsilon_o(h/2\pi)}$$

For $$n=1$$, $$\rightarrow v_1=2.18\times 10^6m/s$$
For $$n=2$$, $$\rightarrow v_1=1.09\times 10^6m/s$$
For $$n=3$$, $$\rightarrow v_1=7.27\times 10^5m/s$$

(b) Orbital perios is given by $$T=\dfrac{2\pi r}{v}; v=velocity$$
Radius, $$r=\dfrac{n^2h^2\epsilon_o}{\pi me^2}$$
So,
$$T_1=1.527\times 10^{-16},\ n=1$$ 
$$T_2=1.22\times 10^{-15},\ n=2$$ 
$$T_3=4.12\times 10^{-15},\ n=3$$ 


In a Geiger Marsden experiment, calculate the distance of closest approach to the nucleus of $$Z=75$$, when an $$\alpha$$-particle of $$5 MeV$$ energy impinges on it before it comes momentarily to rest and reverse its direction.
How will the distance of closest approach be affected when the kinetic energy of the $$\alpha$$-particle is doubled?

If $$r_0$$ be the distance of closest approach between the nucleus and the alpha particle, then the kinetic energy of the alpha particle is $$E_k=\dfrac{1}{4\pi\epsilon_0} \dfrac{(Ze)(2e)}{r_0}$$
Thus, $$r_0 \propto \dfrac{1}{E_k}$$
Hence, when the kinetic energy of alpha particle is doubled, the closet distance $$r_0$$ will be halved.


Find the frequency of revolution of an electron in Bohr's $$2$$nd orbit; if the radius and speed of electron in that orbit is $$2.14\times 10^{-10}$$m and $$1.09\times 10^6m/s$$ respectively. $$[\pi =3.142]$$.

Radius of $$2$$nd orbit $$r = 2.14\times 10^{-10}$$ m
Speed of electron in $$2$$nd orbit $$v = 1.09\times 10^6  m/s$$
Frequency of revolution of an electron in that orbit  $$\nu = \dfrac{v}{2\pi r}$$
$$\therefore$$ $$\nu = \dfrac{1.09\times 10^{6}}{2(3.142)(2.14\times 10^{-10})}$$
$$\implies$$ $$\nu = 8.1\times 10^{14}$$ Hz


An electron in an atom revolved around the nulceus in an orbit of radius $$0.53\overset{o}{A}$$. If the frequency of revolution of an electron is $$9\times 10^9$$ MHz, calculate the orbital angular momentum. [Given: Charge on an electron $$=1.6\times 10^{-19}$$C; Gyromagnetic ratio $$=8.8\times 10^{10}C/kg; \pi =3.142$$].

Given : $$r = 0.53$$ $$A^o$$  and  $$\nu = 9\times 10^{15}$$ Hz  
Gyromagnetic ratio $$\gamma = 8.8\times 10^{10}C/kg$$
Magnetic moment  $$\mu = e\nu (\pi r^2)$$
$$\therefore$$ $$\mu = (1.6\times 10^{-19}) \times (9\times 10^{15}) \times \pi \times (0.53\times 10^{-10})^2$$
$$\implies$$ $$\mu = 12.7\times 10^{-24}$$ $$A m^2$$
Orbital angular momentum $$L = \dfrac{\mu}{\gamma}$$
$$\therefore$$ $$L = \dfrac{12.7\times 10^{-24}}{8.8 \times 10^{10}} = 1.44 \times 10^{-34}$$ $$kg m^3 /s$$


The energy of a $$K-$$electron in tungsten is $$-20\ KeV$$ and of an $$L-$$electron is $$-2\ KeV$$. Find the wavelength of $$X-$$ rays emitted when there is electron jump from $$L$$ to $$K$$ shell.


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When an electron of charge -e and position of charge +e are brought close to each other,they annhilate each other to produce chargeless $$\gamma $$ photons.

This statement is true$$.$$ When $${e^ - }$$ and $${e^ + }$$ annhilate two $$\gamma $$ rays are produced$$.$$


Explain the concept of photon and write its properties.

Concept of Photon:
In $$1990$$, Max Planck proposed his Quantum Theory of radiation. According to this theory, the energy of an electromagnetic wave is not continuously distributed over the wavefront (like the energy of water waves), instead of an electromagnetic wave travels in the form of discrete packets or bundles of energy called Quanta. One quanta of light radiation is called a photon, which travels with the speed of light. 

We can summarize the important properties of photon or the photon picture of the electromagnetic radiation as follows :

1. In the interaction of radiation with matter, radiation behaves as it is made up of particles, called photons.
2. All the photons are emitted by any source which are travelled with speed of light in a free space $$ [c = 3 \times 10^8 \,m/sec] $$ 
3.Each photon has a definite energy depending upon the frequency y of the incident radiation and not on its intensity. 
Energy of each photon, $$ E = hv = \dfrac{hc}{\lambda} $$ 
Where $$h$$ is Planck’s constant and its value is $$ 6.63 \times10^{-34}\, Js$$ or $$ 4.1 × 10^{-15}\, eVs $$  
4. A photon may collide with a material particle. The total energy and the total momentum remain conserved in such a collision. However, the number of photons may not be conserved in a collision. The photon may be absorbed or a new photon may be created.
5. Photons are electrically neutral and are not deflected by electric and magnetic fields.
6. If the intensity of light of a given wavelength is increased, there is an increase in the number of photons emitted by the incident radiations on a given area in a given time. But the energy of each photon remains the same.
7. The rest mass of a photon is consider to be zero.


Answer in brief.
Show that the frequency of the first line in the Lyman series is equal to the difference between the limiting frequencies of Lyman and the Balmer series.

The 'series limit' refers to the 'shortest wavelength'.
$$f= k \left( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right)$$
Here, $$k$$ is a constant.
For the series limit of Lyman series,
$$n_1=1$$
$$n_2 = \infty$$
For the series limit of Balmer series,
$$n_1=2$$
$$n_2 = \infty$$

Thus, the difference in the frequencies of the series of Lyman series and Balmer series is equal to the frequency of the first line of the Lyman series.


Determine the series limit of Balmer, Paschen, and Pfund series, given the limit for Lyman series is $$912 \overset{\circ}{A}$$

Series Limit of Balmer series
$$n=1$$

$$\dfrac{1}{\lambda}= R \left( \dfrac{1}{n^2} \right) = R$$

Given: $$\dfrac{1}{R}= 912 \overset{\circ}{A}$$

Series Limit of Balmer series
$$n=2$$

$$\dfrac{1}{\lambda}= R \left( \dfrac{1}{n^2} \right) = \dfrac{R}{4}$$

$$\dfrac{1}{\lambda}= R_H \left( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right) $$


Suppose the potential energy between an electron and a proton at a distance $$r$$ is given by $$-ke^2/3r^3$$. Use Bohr's theory to obtain energy level of such a hypothetical atom.

Given, potential energy, $$U=-\dfrac{ke^2}{3r^3}$$
The electrostatic force between electron and proton at a distance r is given by, $$F=\dfrac{dU}{dr}=\dfrac{ke^2}{r^4}$$
According to Bohr's first postulate, $$\dfrac{mv^2}{r}=F=\dfrac{ke^2}{r^4}   ...(1)$$
According to Bohr's second postulate, $$mvr=\dfrac{nh}{2\pi}  ........(2)$$
From (1) and (2), $$r=\dfrac{4\pi^2e^2km}{n^2h^2}$$
Total energy of the electron, $$E=K+U=\dfrac{1}{2}mv^2-\dfrac{ke^2}{3r^3}=\dfrac{ke^2}{2r^3}-\dfrac{ke^2}{3r^3}=\dfrac{ke^2}{6r^3}$$  using (1)
Substituting the value of $$r$$, we get $$E=\dfrac{n^6h^6}{6(2\pi)^6m^3k^2e^4}$$


Answer the following question, which help you understand the difference between Thomson's model and Rutherford's model better.
Is the average angle of deflection of $$\alpha$$-particle by a thin gold foil predicted by Thomson's model much less, about the same, or much greater than that predicted by Rutherford's model?

Thomson's atom model cannot explain the scattering of $$\alpha$$-particles through large angles and particularly through $$180^{0}$$. It is because, the light particles like electrons cannot deflect the massive a-particles through $$180^{0}$$.


What kind of Zeeman effect, normal or anomalous, is observed in a weak magnetic field in the case of spectral lines caused by the following transitions:
(a) $$^{1}P \rightarrow ^{1}S $$;     (b) $$^{2}D_{5/2} \rightarrow ^{2}P_{3/2} $$;           (c) $$^{3}D_{1} \rightarrow ^{3}P_{0}$$;           (d) $$^{5}I_{5} \rightarrow ^{5}H_{4} $$ ?


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Class 12 Engineering Physics Extra Questions