Atoms And Nuclei MCQ Questions & Answers in Modern Physics | Physics
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81.
As an electron makes a transition from an excited state to the ground state of a hydrogen - like atom/ion :
A
kinetic energy decreases, potential energy increases but total energy remains same
B
kinetic energy and total energy decrease but potential energy increases
C
its kinetic energy increases but potential energy and total energy decrease
D
kinetic energy, potential energy and total energy decrease
Answer :
its kinetic energy increases but potential energy and total energy decrease
$$\eqalign{
& U = - K\frac{{z{e^2}}}{r};T.E = - \frac{k}{2}\frac{{z{e^2}}}{r} \cr
& K.E = \frac{k}{2}\frac{{z{e^2}}}{r}. \cr} $$
Here $$r$$ decreases
82.
Hydrogen atom excites energy level from fundamental state to $$n = 3.$$ Number of spectral lines according to Bohr, is
83.
Which of the following statements are true regarding Bohr’s model of hydrogen atom?
(I) Orbiting speed of electron decreases as it shifts to discrete orbits away from the nucleus
(II) Radii of allowed orbits of electron are proportional to the principal quantum number
(III) Frequency with which electrons orbit around the nucleus in discrete orbits is inversely proportional to the cube of principal quantum number
(IV) Binding force with which the electron is bound to the nucleus increases as it shifts to outer orbits
Select correct answer using the codes given below. Codes :
A
I and II
B
II and IV
C
I, II and III
D
II, III and IV
Answer :
I and II
Orbital speed varies inversely as the radius of the orbit.
$$v \propto \frac{1}{n}$$
84.
Hydrogen atoms are excited from ground state of the principal quantum number 4. Then, the number of spectral lines observed will be
A
3
B
6
C
5
D
2
Answer :
6
Number of spectral lines observed in hydrogen spectrum is given by
$$\eqalign{
& = \frac{{n\left( {n - 1} \right)}}{2} \cr
& = \frac{{4\left( {4 - 1} \right)}}{2} \cr
& = 6 \cr} $$
Where, $$n$$ = principal quantum number = number of orbits.
85.
Energy required for the electron excitation in $$L{i^{ + + }}$$ from the first to the third Bohr orbit is :
A
$$36.3\,eV$$
B
$$108.8\,eV$$
C
$$122.4\,eV$$
D
$$12.1\,eV$$
Answer :
$$108.8\,eV$$
Energy of excitation,
$$\eqalign{
& \Delta E = 13.6{Z^2}\left( {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right)eV \cr
& \Rightarrow \Delta E = 13.6{\left( 3 \right)^2}\left( {\frac{1}{{{1^2}}} - \frac{1}{{{3^2}}}} \right) = 108.8\,eV \cr} $$
86.
A proton and an alpha particle both enter a region of uniform magnetic field $$B,$$ moving at right angles to the field $$B.$$ If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is $$1\,MeV,$$ the energy acquired by the alpha particle will be
A
$$4\,MeV$$
B
$$0.5\,MeV$$
C
$$1.5\,MeV$$
D
$$1\,MeV$$
Answer :
$$1\,MeV$$
Radius in magnetic fields of circular orbit,
$$R = \frac{{mV}}{{qB}} = \frac{{\sqrt {2mE} }}{{qB}}$$
and total energy of a moving particle in a circular orbit, $$E = \frac{{{q^2}{B^2}{R^2}}}{{2m}}$$
For a proton enter in a region of magnetic field,
$${E_1} = \frac{{{e^2} \times {B^2} \times {R^2}}}{{2 \times {m_p}}}\,.......\left( {\text{i}} \right)$$
where $${{m_p}}$$ is the mass of proton.
Similarly, for a $$\alpha $$-particle moves in a uniform magnetic field
$${E_2} = \frac{{{{\left( {2e} \right)}^2} \times {B^2} \times {R^2}}}{{2 \times \left( {4{m_p}} \right)}}\,\,\left[ {\because {m_\alpha } = 4{m_p}} \right]\,.......\left( {{\text{ii}}} \right)$$
Dividing Eq. (ii) by Eq. (i), we get
$$\eqalign{
& \frac{{{E_2}}}{{{E_1}}} = \frac{{{{\left( {2e} \right)}^2} \times {B^2} \times {R^2}}}{{2 \times \left( {4{m_p}} \right)}} \times \frac{{2 \times {m_p}}}{{{e^2} \times {B^2} \times {R^2}}} \cr
& \frac{{{E_2}}}{{{E_1}}} = 1 \Rightarrow {E_2} = {E_1} = 1\,MeV \cr} $$
87.
An electron in a hydrogen atom makes a transition from first excited state to ground state. The equivalent current due to circulating electron -
A
increases 2 times
B
increases 4 times
C
increases 8 times
D
remains the same
Answer :
increases 8 times
$$\eqalign{
& i = \frac{q}{T} \cr
& {\text{Now,}}\,\,{T^2} \propto {r^3} \propto {n^6} \cr
& \Rightarrow i \propto \frac{1}{{{n^3}}} \cr} $$
88.
The transition from the state $$n = 4$$ to $$n = 3$$ in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from :
A
$$3 \to 2$$
B
$$4 \to 2$$
C
$$5 \to 4$$
D
$$2 \to 1$$
Answer :
$$5 \to 4$$
It is given that transition from the state $$n = 4$$ to $$n = 3$$ in a hydrogen like atom result in ultraviolet radiation. For infrared radiation the energy gap should be less. The only option is $$5 \to 4.$$
89.
When electron jumps from $$n = 4$$ to $$n =2$$ orbit, we get
A
second line of Lyman series
B
second line of Balmer series
C
second line of Paschen series
D
an absorption line of Balmer series
Answer :
second line of Balmer series
(A) Second line of Lyman series corresponds to the transition $$n = 3 \to n = 1$$
(B) Second line of Balmer series corresponds to the transition $$n = 4 \to n = 2$$
(C) Second line of Paschen series corresponds to the transition $$n = 5 \to n = 3$$
(D) An absorption line of Balmer series arises when electron jumps from $$n =2$$ to any other higher state.
Thus, choice (B) is correct. NOTE
For Lyman series
$${n_2} = 2,3,4......... \to {n_1} = 1$$
For Balmer series
$${n_2} = 3,4,5......... \to {n_1} = 2$$
For Paschen Series
$${n_2} = 4,5,6......... \to {n_1} = 3$$
For Brackett series
$${n_2} = 5,6,7......... \to {n_1} = 4$$
For Pfund series
$${n_2} = 6,7,8......... \to {n_1} = 5$$
90.
The element which has a $${K_\alpha }$$ x-rays line of wavelength $$1.8\,\mathop {\text{A}}\limits^ \circ $$ is ($$R = 1.1 \times {10^{ - 7}}{m^{ - 1}},b = 1$$ and $$\sqrt {\frac{5}{{33}}} = 0.39$$ )