Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 11

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Weekly Quiz Competition

Question 1
1 / -0

The de - Broglie wavelength associated with the electron in the $$n=4$$ level is :

A
half of the de-Broglie wavelength of the electron in the ground state

B
four times the de-Broglie wavelength of the electron in the ground state

C
$$1/4^{th}$$ of the de-Broglie wavelength of the electron in the ground state

D
two times the de-Broglie wavelength of the electron in the ground state

Solution

De-Broglie wavelength is given as

$$\lambda =\dfrac{h}{mv}$$...(i) where h is planck constant

The velocity of electron in Bohr model is given as

$$v=\dfrac{h}{2\pi m_e a_\circ{} n}$$...(i)

in the ground state n=1

$$v_{n=1}=\dfrac{h}{2\pi m_e a_\circ{} }$$

$$v_{n=4}=\dfrac{h}{2\pi m_e a_\circ{} 4}$$

$$v_{n=4}=\dfrac{v_{n=1}}{4}$$...(iii)

from equation (i)

$$\lambda_{n=4}=\dfrac{h}{m_e v_{n=4} }$$

substituting value of $$v_{n=4}$$ from equation (iii)

$$\lambda_{n=4}=\dfrac{h}{m_e \dfrac{ v_{n=1}}{4} }$$

$$\lambda_{n=4}=4 \lambda_{n=1}$$

Hence correct answer will be option B.
Question 2
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A particle $$A$$ of mass $$m$$ and initial velocity $$v$$ collides with a particle $$B$$ of mass $$\dfrac{m}{2}$$ which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths $${\lambda}_A$$ to $${\lambda}_B$$ after the collision is :

A
$$\dfrac{{\lambda}_A}{{\lambda}_B}=\dfrac{1}{2}$$

B
$$\dfrac{{\lambda}_A}{{\lambda}_B}=\dfrac{1}{3}$$

C
$$\dfrac{{\lambda}_A}{{\lambda}_B}=2$$

D
$$\dfrac{{\lambda}_A}{{\lambda}_B}=\dfrac{2}{3}$$

Solution

From conservation of linear momentum, $$mv= mv_B + \dfrac{m}{2} v_B$$

$$v= v_A+ \dfrac{v_B}{2}$$

Also $$ v^2 = v_a^2 + \dfrac{v_B^2}{2}$$

$${(v_A+ \dfrac{v_B}{2})}^2= v_A^2 + {(\dfrac{v_B}{2})}^2$$

$$v_Av_B = \dfrac{v_B}{4}$$

$$v_B= 4 v_A$$ and $$m_B= \dfrac{m_A}{2}$$

$$P_A=m_Av_A$$

$$P_B= \dfrac{m_A}{2} \times 4 v_A= 2 P_A$$

$$\dfrac{\lambda_A}{\lambda_B}= \dfrac{P_B}{P_A}= \dfrac{2 P_A}{P_A}= 2$$
Question 3
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For which of the following particles will it be most difficult to experimentally verify the de-Broglie relationship?

A
An electron

B
A proton

C
An $$\alpha-particle$$

D
A dust particle

Solution

A dust particle will it be most difficult to experimentally verify the de-Broglie relationship .
Question 4
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If electron charge e, electron mass m, speed of light in vacuum c and Planck's constant h are taken as fundamental constant h are taken as fundamental quantities, the permeability of vacuum $$\mu_0$$ can be expressed in units of

A
$$\left (\dfrac {mc^2}{he^2}\right )$$

B
$$\left (\dfrac {h}{me^2}\right )$$

C
$$\left (\dfrac {hc}{me^2}\right )$$

D
$$\left (\dfrac {h}{ce^2}\right )$$

Solution

The fine structure constant is defined as $$\alpha = \dfrac{e^2}{4\pi \epsilon _0 \hbar c}$$ and is dimensionless where $$\hbar = \dfrac{h}{2\pi}$$

$$ \therefore \alpha = \dfrac{e^2 c}{4\pi \epsilon _0 \hbar c^2}=\dfrac{e^2 c \epsilon _0 \mu _0}{4\pi \epsilon _0 \hbar}$$ where $$c = \dfrac{1}{\sqrt{\epsilon _0 \mu _0}}$$

$$ \Rightarrow \mu _0 = \dfrac{ \hbar}{e^2 c}$$
Question 5
1 / -0

De-Broglie wavelength of an electron accelerated by a voltage of 50 V is close to: $$(|e|=1.6 \times 10^{-19} C, m_e=9.1 \times 10^{-31}kg, h=6.6 \times 10^{-34} Js).$$

A
$$0.5 \mathring {A} $$

B
$$1.7 \mathring {A} $$

C
$$2.4 \mathring {A} $$

D
$$1.2 \mathring {A} $$

Solution

$$\lambda = \dfrac{h}{p}= \dfrac{h}{\sqrt{2mE}}= \dfrac{h}{\sqrt{2mqV}}$$
$$\lambda = \dfrac{ 6.6 \times 10 ^ {-34} }{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 50}} = 1.72 \times 10^{-10} = 1.72 A$$
Option B
Question 6
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Photon of frequency $$'v\ '$$ has a momentum associated with it. If $$'c\ '$$ is the velocity of light, the momentum is

A
$$v/c$$

B
$$hvc$$

C
$$hv/c$$$$^{2}$$

D
$$hv/c$$

Solution

We know that energy of photon is $$E=h\nu ....(1)$$
Momentum of photo , $$p=mc=mc^2/c=E/c$$ where $$E=mc^2$$
using (1), $$p=h\nu/c$$
Question 7
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A parallel beam of electrons travelling in x-direction falls on a slit of width d. If after passing the slit, an electron acquires momentum $$p_y$$ in the y direction, then for a majority of electrons passing through the slit (h is Planck's constant).

A
$$|P_y|d < h$$

B
$$|P_y|d \simeq h$$

C
$$|P_y|d >> h$$

D
$$|P_y|d > h$$

Solution

From Heisenberg's uncertainty principle, $$ \Delta p \Delta y \geq h$$

For most electrons, $$\Delta p < p_y$$ (change in momentum is small )

and $$\Delta y < d$$ (most electrons lie in central fringe)

$$ \Rightarrow |p_y|d \geq h$$

Minimum value of $$ \Rightarrow |p_y|d$$ is $$h$$.
Question 8
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Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de Broglie wavelengths are $${ \lambda }_{ 1 }$$ and $${ \lambda }_{ 2 }$$, their de Broglie wavelength in the frame of reference attached to their centre of mass is:

A
$${ \lambda }_{ CM }={ \lambda }_{ 1 }={ \lambda }_{ 2 }\quad $$

B
$$\cfrac { 1 }{ { \lambda }_{ CM } } =\cfrac { 1 }{ { \lambda }_{ 1 } } +\cfrac { 1 }{ { \lambda }_{ 2 } } $$

C
$${ \lambda }_{ CM }=\cfrac { 2{ \lambda }_{ 1 }{ \lambda }_{ 2 } }{ \sqrt { { { \lambda }_{ 1 } }^{ 2 }+{ { \lambda }_{ 2 } }^{ 2 } } } $$

D
$${ \lambda }_{ CM }=\left( \cfrac { { \lambda }_{ 1 }+{ \lambda }_{ 2 } }{ 2 } \right) $$

Solution

Since $$\lambda = \dfrac{h}{mv}$$
$$v = \dfrac{h}{m \lambda}$$

Let $$v_1$$ and $$v_2$$ are the speeds of electrons

$$v_{cm} = \dfrac{v_1 + v_2}{2}$$
$$\dfrac{h}{2m \lambda_{cm}} = \dfrac{1}{2}( \dfrac{h}{m \lambda_1} + \dfrac{h}{m \lambda_2})$$

$$\dfrac{1}{\lambda_{cm}} = \dfrac{1}{\lambda_1} + \dfrac{1}{\lambda_2}$$
Question 9
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The de-Broglie wavelength $$(\lambda_{B})$$ associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state $$(\lambda_{G})$$ by :

A
$$\lambda_{B} = \lambda_{G}/3$$

B
$$\lambda_{B} = \lambda_{G}/2$$

C
$$\lambda_{B} = 2\lambda_{G}$$

D
$$\lambda_{B} = 3\lambda_{G}$$

Solution

de- Broglie wavelength $$\lambda=\dfrac{h}{mv}$$

By conservation of angular momentum, $$mvr=\dfrac{nh}{2\pi}$$

$$\dfrac{h}{mv}=\dfrac{2\pi r}{n}$$

$$\lambda=\dfrac{2\pi r}{n}$$

$$r=a_{0}\dfrac{n^{2}}{Z}$$

$$\lambda=\dfrac{2\pi a_{0}n}{Z}$$

As the atom is hydrogen $$Z=1$$

$$\lambda_{B}=\dfrac{2\pi a_{0}3}{Z}$$

$$\lambda_{G}=\dfrac{2\pi a_{0}}{Z}$$

$$\lambda_{B}=3\lambda_{G}$$
Question 10
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An electron (mass $$m$$) with initial velocity $$\vec { v } ={ v }_{ 0 }\hat { i } +{ v }_{ 0 }\hat { j } $$ is an electric field $$\vec { E } =-{ E }_{ 0 }\hat { k } $$. If $${ \lambda }_{ 0 }$$ is initial de-Broglie wave length at time $$t$$ is given by

A
$$\cfrac { { \lambda }_{ 0 } }{ \sqrt { 1+\cfrac { { e }^{ 2 }{ E }^{ 2 }{ t }^{ 2 } }{ 2{ m }^{ 2 }{ v }_{ 0 }^{ 2 } } } } $$

B
$$\cfrac { { \lambda }_{ 0 }\sqrt { 2 } }{ \sqrt { 1+\cfrac { { e }^{ 2 }{ E }^{ 2 }{ t }^{ 2 } }{ { m }^{ 2 }{ v }_{ 0 }^{ 2 } } } } $$

C
$$\cfrac { { \lambda }_{ 0 } }{ \sqrt { 2+\cfrac { { e }^{ 2 }{ E }^{ 2 }{ t }^{ 2 } }{ { m }^{ 2 }{ v }_{ 0 }^{ 2 } } } } $$

D
$$\cfrac { { \lambda }_{ 0 } }{ \sqrt { 1+\cfrac { { e }^{ 2 }{ E }_{ 0 }^{ 2 }{ t }^{ 2 } }{ { m }^{ 2 }{ v }_{ 0 }^{ 2 } } } } $$

Solution

Initially
momentum, $$P=\dfrac{h}{\lambda_0}$$ $$\because V_i = \sqrt{V_0^2 + V_0^2} = \sqrt{2}V_0$$

$$m(\sqrt{2}V_0) = \dfrac{h}{\lambda_0}$$

Now, velocity as a function of time $$= V_0\hat{i} + V_0\hat{j} + \dfrac{eE_0}{m} + \hat{k}$$

because only force is acting in $$Z-dir^n$$. So velocity will be change in $$Z.dir^n$$ and keeping constant in $$x-y \ dir^n$$.

$$\therefore V_{net} = \sqrt{V_0^2+V_0^2 + \left(\left( \dfrac{eE_0}{m}\right)\right)^2}$$

$$\Rightarrow P = \dfrac{h}{\lambda}$$

$$\Rightarrow mV_{net} = \dfrac{h}{\lambda}$$

$$\Rightarrow \lambda = \dfrac{h}{mV_{net}} = \dfrac{h}{m\sqrt{2V_0^2 + \dfrac{e^2E_0^2}{m^2}t^2}}$$ $$\left( \because \lambda_0 = \dfrac{h}{m\sqrt{2}V_0}\right)$$

$$\Rightarrow \lambda = \dfrac{h}{\sqrt{\left(1 + \dfrac{e^2E_0^2}{2m^2V_0^2}t^2\right)}}$$

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Dual Nature of Radiation and Matter Test - 11

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Dual Nature of Radiation and Matter Test - 11

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Question 1

half of the de-Broglie wavelength of the electron in the ground state

four times the de-Broglie wavelength of the electron in the ground state

$$1/4^{th}$$ of the de-Broglie wavelength of the electron in the ground state

two times the de-Broglie wavelength of the electron in the ground state

Solution

Question 2

$$\dfrac{{\lambda}_A}{{\lambda}_B}=\dfrac{1}{2}$$

$$\dfrac{{\lambda}_A}{{\lambda}_B}=\dfrac{1}{3}$$

$$\dfrac{{\lambda}_A}{{\lambda}_B}=2$$

$$\dfrac{{\lambda}_A}{{\lambda}_B}=\dfrac{2}{3}$$

Solution

Question 3

An electron

A proton

An $$\alpha-particle$$

A dust particle

Solution

Question 4

$$\left (\dfrac {mc^2}{he^2}\right )$$

$$\left (\dfrac {h}{me^2}\right )$$

$$\left (\dfrac {hc}{me^2}\right )$$

$$\left (\dfrac {h}{ce^2}\right )$$

Solution

Question 5

$$0.5 \mathring {A} $$

$$1.7 \mathring {A} $$

$$2.4 \mathring {A} $$

$$1.2 \mathring {A} $$

Solution

Question 6

$$v/c$$

$$hvc$$

$$hv/c$$$$^{2}$$

$$hv/c$$

Solution

Question 7

$$|P_y|d < h$$

$$|P_y|d \simeq h$$

$$|P_y|d >> h$$

$$|P_y|d > h$$

Solution

Question 8

$${ \lambda }_{ CM }={ \lambda }_{ 1 }={ \lambda }_{ 2 }\quad $$

$$\cfrac { 1 }{ { \lambda }_{ CM } } =\cfrac { 1 }{ { \lambda }_{ 1 } } +\cfrac { 1 }{ { \lambda }_{ 2 } } $$

$${ \lambda }_{ CM }=\cfrac { 2{ \lambda }_{ 1 }{ \lambda }_{ 2 } }{ \sqrt { { { \lambda }_{ 1 } }^{ 2 }+{ { \lambda }_{ 2 } }^{ 2 } } } $$

$${ \lambda }_{ CM }=\left( \cfrac { { \lambda }_{ 1 }+{ \lambda }_{ 2 } }{ 2 } \right) $$

Solution

Question 9

$$\lambda_{B} = \lambda_{G}/3$$

$$\lambda_{B} = \lambda_{G}/2$$

$$\lambda_{B} = 2\lambda_{G}$$

$$\lambda_{B} = 3\lambda_{G}$$

Solution

Question 10

$$\cfrac { { \lambda }_{ 0 } }{ \sqrt { 1+\cfrac { { e }^{ 2 }{ E }^{ 2 }{ t }^{ 2 } }{ 2{ m }^{ 2 }{ v }_{ 0 }^{ 2 } } } } $$

$$\cfrac { { \lambda }_{ 0 }\sqrt { 2 } }{ \sqrt { 1+\cfrac { { e }^{ 2 }{ E }^{ 2 }{ t }^{ 2 } }{ { m }^{ 2 }{ v }_{ 0 }^{ 2 } } } } $$

$$\cfrac { { \lambda }_{ 0 } }{ \sqrt { 2+\cfrac { { e }^{ 2 }{ E }^{ 2 }{ t }^{ 2 } }{ { m }^{ 2 }{ v }_{ 0 }^{ 2 } } } } $$

$$\cfrac { { \lambda }_{ 0 } }{ \sqrt { 1+\cfrac { { e }^{ 2 }{ E }_{ 0 }^{ 2 }{ t }^{ 2 } }{ { m }^{ 2 }{ v }_{ 0 }^{ 2 } } } } $$

Solution

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