3 Marks Question

Question 13 Marks

Explain the assumption related to quantization of angular momentum in Bohr's theory.###Explain quantum condition for orbital motion of electron in hydrogen atom.###Explain Bohr's second postulate of quantisation by de Broglie hypothesis.

Answer

Bohr's Second Postulate : Electron revolves only those orbits in which the angular momentum $( L =m v r)$ is multiple integer of $\frac{h}{2 \pi}$.
According to Bohr's this postulate, $L =\frac{n h}{2 \pi}$
or $\quad m v r=\frac{n h}{2 \pi} \quad$ Where $n$ is an integer whose values are $n=1,2,3, \ldots \ldots$ respectively.
$n$ is principal quantum number and this condition is called Bohr's quantum condition. Motion of electron is limited in possible orbits by this condition.
The stable orbits of different radii of electron can be obtained by given the different values to $n$.

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Question 23 Marks

Calculate the distance of closest approach of $\alpha-$ particle which has kinetic energy $4.5 MeV,$ collides with the nucleus of $Z=80$ and reversed in its direction after rest.

Answer

Let distance of closest approach is $ ' r \ '$
$K =4.5 MeV$
$=4.5 \times 10^6 eV$
$Z =80$
$K =\frac{1}{4 \pi \epsilon_0} \cdot \frac{(Ze)(2 e)}{r}$
or $ r=\frac{1}{4 \pi \epsilon_0} \cdot \frac{(Ze)(2 e)}{k}$
$=\frac{9 \times 10^9 \times 2 \times 80 e^2}{4.5 MeV}$
$=\frac{9 \times 10^9 \times 2 \times 80 \times 1.6 \times 10^{-19}}{4.5 \times 10^6}$
$=\frac{9 \times 160 \times 1.6}{4.5} \times 10^{-16}$
$=512 \times 10^{-16} m$
$=5.12 \times 10^{-14}m.$

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Question 33 Marks

An $\alpha -$ particle collides with nucleus after passing through the potential $V$ volt. Prove that distance of closest approach of particle of atomic number $Z$ to the nucleus will be $14.4( Z / V ) \ \mathring A $.
$($Given that : $1 / 4 \pi \varepsilon_0=9.0 \times 10^9$ Newton $-$ meter $^2 /$ Coulomb $^2$ and $e=1.6 \times 10^{-19}$ Coulomb$)$

Answer

Energy of the $\alpha-$ particle to reach the closest to the nucleus :
Kinetic energy of $\alpha-$ particle $=$ Electrostatic potential energy of $\alpha-$ particle and nucleus system.
$2 e \times V=\frac{1}{4 \pi \in_0}\left(\frac{2 e \times Z e }{r_0}\right)$
$=9 \times 10^9\left(\frac{2 e \times Z e }{r_0}\right)$
$\Rightarrow 2 eV=\frac{2 \times 9 \times 10^9 \times Z e^2}{r_0} $
$\Rightarrow r_0=\frac{9 \times 10^9 \times Z e }{V} $ meter
Putting the values $=9 \times 10^9 \times 1.6 \times 10^{-19}\left(\frac{Z}{V}\right)$ meter
$=14.4 \times 10^{-10}\left(\frac{Z}{V}\right) $ meter
$=\left[14.4\left(\frac{Z}{V}\right)\right] \ \mathring A $

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Question 43 Marks

How much is the minimum energy of electron of excited hydrogen atom so that three spectral lines of hydrogen spectrum could be obtained?

Answer

To obtain the three spectral lines in hydrogen spectrum it is necessary that electron have that amount of energy, which excites the atom, from $n=1$ to $n=3,$ so that three transitions
$=(3 \rightarrow 2,3 \rightarrow 1,2 \rightarrow 1) $ will be obtained.
In this condition from the formula $, E _n=-\frac{13.6}{n^2} eV$
Therefore,
$ E _1=-\left(\frac{13.6}{1}\right) eV =-13.6 eV$
$E _3=\left(\frac{13.6}{3^2}\right) eV =-1.5 eV $
Required min. energy $= E _3- E _1$
$=-1.5 eV -(-13.6 eV )$
$=-1.5 eV +13.5 eV$
$=12.1 eV $
Although to obtain the observed three spectral lines of atom $(1 \rightarrow 2,1 \rightarrow 3,1 \rightarrow 4)$, more energy $\left(E_4-E_1\right)=12.75eV$ should be required comparatively that of above.

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Question 53 Marks

Energy of electron in hydrogen atom expressed by the following formula :
$E _n=-\left(\frac{13.6 eV}{n^2}\right)$ ; Where $, n=1,2,3, \ldots$
by using this formula, proved that :
$(a)$ The energy of electron $-6.8 eV$ cannot be in $H -$ atom.
$(b)$ The distance of attached lines in spectrum of hydrogen.

Answer

Formula, $ E _n=-\left(\frac{13.6 eV }{n^2}\right)$, solving it by putting $n=1,2,3, \ldots \ldots \infty,$
$E_1=-\left(\frac{13.6}{1^2}\right) eV$
$=-13.6 eV$
$E_2=-\left(\frac{13.6}{2^2}\right) eV$
$=-\frac{13.6}{4} eV=-3.4 eV$
$E_3=-\left(\frac{13.6}{3^2}\right) eV$
$=-\frac{13.6}{9} eV=-1.51 eV$
$E_4=-\left(\frac{13.6}{4^2}\right) eV$
$=-\frac{13.6}{16} eV=-0.85 eV$
$E_{\infty}=-\left(\frac{13.6}{\infty}\right)=0 eV$
$(a)$ Therefore, it is clear that energy of electron in hydrogen atom cannot be $-6.8 eV$.
$(b)$ Also be clear that, on increasing the $n$, difference of two attached lines $($energy levels$)$ decreases.
Therefore, the distance between them decreases.

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Question 63 Marks

The radius of innertmost orbit of hydrogen atom is $5.3 \times 10^{-11} m$. What is the radius of second excited state?

Answer

Radius of $n^{\text {th }}$ orbit of hydrogen atom :
$r_n=n^2 r_1$
According to question :
$r_1=5.3 \times 10^{-11} m$
Here given that : $n=3$
$\because$ For second excited state $n=3$
So, $r_3=(3)^2 \times 5.3 \times 10^{-11}$
$=9 \times 5.3 \times 10^{-11}=47.7 \times 10^{-11} m$

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Question 73 Marks

Explain the shortcomings of Bohr's Theory.

Answer

Shortcomings of Bohr's Theory :
(i) Bohr's Theory explains the spectra of one-electron atoms only such as hydrogen, singly-ionised helium, doublyionised lithium, etc. It fails to explain the spectra of other atoms.
(ii) In this theory, nucleus assumed as stable, but it is possible only when mass of nucleus is infinite.
(iii) In this theory, orbits of electrons assumed circular while almost these are elliptical.
(iv) Bohr's theory does not explain about the intensities of spectral lines.
(v) On the basis of this theory quantization of angular momentum has no logical fact.
(vi) This theory does not explain the 'fine structure' of spectral lines.
(vii) The theory cannot fully explain the splitting of a spectral line into a number of components under the effect of a magnetic field. This effect is called Zeeman Effect, which is not explained by Bohr's theory In this way splitting of a spectral line under the effect of an electric field which is called Stark effect.

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Question 83 Marks

In $H _2$ atom, $r_0$ is radius of first Bohr's orbit. What will be the radius of second orbit? What will be the radius of single ionized helium atom?

Answer

$\because r \propto n^2$ for $H _2-$atom, therefore radius of its second orbit :
$r_2=2^2 r_1=2^2 \times r_0=4 r_0$
$r \propto\left(\frac{n^2}{Z}\right)$ is similar to $H _2-$atom and $Z =2$ for single ionised helium atom
$=\frac{\text { Radius of second orbit for single ionised He atom }}{\text { Radius of second orbit of } H_2 \text { atom }}$
$=\frac{\frac{2^2}{2}}{2^2}=\frac{1}{2}$
Therefore, Radius of second orbit for single ionised $He-$atom
$=\frac{\text { Radius of second orbit } H_2-\text { atom }}{2}$
$=\frac{4 r_0}{2}=2 r_0$

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Question 93 Marks

For given time interval. Show the graphical representation at different angles for some $\alpha$-particle scattering.

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Question 103 Marks

In hydrogen atom, there is single electron only but many lines in its emission spectrum. How it does? Explain the reason.

Answer

Each atom have some definite energy levels. In normal state electron of hydrogen atom remains in low energy level. When atom gets energy from outside, electron jumps into higher energy level and leave the minimum energy level i.e. it becomes excited. Electron leave the higher energy level within $10^{-8} sec$. Now it can return in lowest energy level or can be return passing through the lower energy level to lowest one. Since, any light source (hydrogen lamp) has large numbers. of atoms therefore all possible transition initiates in source and many lines has seen in spectrum.

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Question 113 Marks

In accordance with the Bohr's model, find the quantum that characterises the earth's revolution around the sun in an orbit of radius $1.5 \times 10^{11} m$ with orbital speed $3 \times 10^4 m / s$. (Mass of earth $=6.0 \times 10^{24} kg$. )

Answer

Given that : $r=1.5 \times 10^{11} m$
$v=3 \times 10^4 m / sec$
Mass of earth $m_e=6.0 \times 10^{24} kg$
Apply Bohr's postulates,
$m v r=\frac{n h}{2 \pi}$
or$\quad$$n=\frac{2 \pi m v r}{h}$
Put the values:
$n=\frac{2 \times 3.14 \times 6 \times 10^{24} \times 3 \times 10^4 \times 1.5 \times 10^{11}}{6.63 \times 10^{-34}}$
$=\frac{6.28 \times 27 \times 10^{39}}{6.63 \times 10^{-34}}=2.557 \times 10^{74}$
$\simeq 2.6 \times 10^{74}$

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Question 123 Marks

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.

Answer

Here, hydrogen atom absorbs one photon, which makes it excited upto n = 4. Therefore values of wavelength of photon will:
$\frac{1}{\lambda}=R\left(\frac{1}{1^2}-\frac{1}{4^2}\right)=R\left(1-\frac{1}{16}\right)=\frac{15}{16} R$
Or wavelength $\lambda=\frac{16}{15} R=\frac{16}{15 \times 1.09 \times 10^7}$
$=0.9724 \times 10^{-7} m$
$=97.24 nm$.
We know that, $c=v \lambda$
$\therefore$ Frequency, $v=\frac{c}{\lambda}=\frac{3 \times 10^8}{97.24 \times 10^{-9}}$
$=3.1 \times 10^{15} Hz$

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Question 133 Marks

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

Answer

Potential energy of electron,
$E _p=-\frac{1}{4 \pi \in_0} \frac{e^2}{r}$
and Kinetic energy, $E _k=\frac{1}{2} \times \frac{1}{4 \pi \in_0} \frac{e^2}{r}=-\frac{1}{2} E _p$
Given, $E _p+ E _k=-13.6 eV$
$\because$ Total energy in ground state, $E =-13.6 eV$
or$\quad$$E _p=-\frac{1}{2} E _p=-13.6 eV$
or$\quad$$\frac{1}{2} E _p=-13.6 eV$
or$E _p=-27.2 eV$
$\therefore \quad E _k=-\frac{1}{2} E _p=\frac{27.2}{2}=13.6 eV$

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