Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 13

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Question 1
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A photon of energy $$10.2 eV$$ collides inelastically with a Hydrogen atom in ground state. After a certain time interval of few micro seconds another photon of energy $$15.0 eV$$ collides inelastically with the same hydrogen atom, then the observation made by a suitable detector is

A
1 photon with energy $$10.2 eV$$ and an electron with energy $$1.4 eV$$

B
2 photon with energy $$10.2 eV$$

C
2 photon with energy $$1.4 eV$$

D
One photon with energy $$3.4 eV$$ and 1 electron with energy $$1.4 eV$$

Solution

$$10.2 eV$$ photon on collision will excite H-atom to first excited state but Hydrogen atom will return to ground state before next collision. Second photon will provide ionization energy to Hydrogen atom, i.e., electron will be ejected with energy = $$1.4 eV$$
Question 2
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Directions For Questions

When a particle is restricted to move along x-axis between $$x = 0$$ and $$x = a$$, where a is of nanometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region,correspond to the formation of standing waves with nodes at its ends $$x=0$$ and $$x=a$$. The wavelength of this standing wave is related to the linear momentum p of the particle according to the de Broglie relation. The energy of the particle of mass m is related to its linear momentum as $$E = p^{2}/2m$$. Thus, the energy of the particle can be denoted by a quantum number n taking values $$1, 2, 3, . $$($$n = 1$$, called the ground state) corresponding to the number of loops in the standing wave.
Use the model described above to answer the following three questions for a particle moving in the line $$x = 0$$ to $$x = a$$. Take $$h = 6.6 \times 10^{-34} Js$$ and $$e = 1.6 \times 10^{-19} C.$$

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The allowed energy for the particle for a particular value of n is proportional to :

A
$$\mathrm{a}^{-2}$$

B
$$\mathrm{a}^{-3/2}$$

C
$$\mathrm{a}^{-1}$$

D
$$\mathrm{a}^{2}$$

Solution

$$ a = \cfrac{n \lambda}{2}$$
De broglie wavelength
$$ \lambda = \cfrac{h}{mv}$$
Subsitute
$$ p =\cfrac{nh}{2a}$$
$$ E = \cfrac{1}{2}mv^2= \cfrac{p^2}{2m} =\cfrac{n^2h^2}{8a^2m}$$
Question 3
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Light of wavelength $$\displaystyle { \lambda }_{ ph }$$ falls on a cathode plate inside a vacuum tube as shown in the figure. The work function of the cathode surface is $$\displaystyle \phi $$ and the anode is a wire mesh of conducting material kept at a distance d from the cathode. A potential difference V is maintained between the electrodes. If the minimum de Broglie wavelength of the electrons passing through the anode is $$\displaystyle { \lambda }_{ e }$$, which of the following statement(s) is(are) true?

A
$$\displaystyle { \lambda }_{ e }$$ increases at the same rate as $${ \lambda }_{ ph }$$ for $$\lambda_{ph} < hc / \phi$$

B
$$\displaystyle { \lambda }_{ e }$$ is approximately halved, if d is doubled

C
$$\displaystyle { \lambda }_{ e }$$ decreases with increase in $$\displaystyle \phi $$ and $$\displaystyle { \lambda }_{ ph }$$

D
For large potential difference $$\displaystyle \left( V>>\phi /e \right) { \lambda }_{ e }$$ is approximately halved if V is made four time

Solution

$${ \lambda }_{ e }=$$ minimum be broglie wavelength.

$$\dfrac { hc }{ { \lambda }_{ ph } } =\dfrac { hc }{ { \lambda }_{ e } } +\phi +eV$$

$$\Rightarrow$$ $$hc(\dfrac { -1 }{ { \lambda }_{ ph }^{ 2 } } d{ \lambda }_{ ph })=hc(\dfrac { -1 }{ { \lambda }_{ e }^{ 2 } } d{ \lambda }_{ e })$$

$$\Rightarrow$$ $$\dfrac { d{ \lambda }_{ ph } }{ d{ \lambda }_{ e } } =\dfrac { { \lambda }_{ ph }^{ 2 } }{ { \lambda }_{ e }^{ 2 } } $$

This denies option A, B and C.

$$V>>\dfrac { \phi }{ e } $$

Energy of electron,

$$E=\dfrac { { P }^{ 2 } }{ 2m } $$

$$\lambda_e =\dfrac { h }{ p } =\dfrac { h }{ \sqrt { 2mE } } $$$$=\dfrac{h}{\sqrt{2meV}}$$

This supports option D.
Question 4
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Light of wavelength $$500\ nm$$ is incident on a metal with a work function $$2.28\ eV$$. The de Borglie wavelength of the emitted electron is:

A
$$\leq 2.8 \times 10^{-12}m$$

B
$$< 2.8 \times 10^{-10}m$$

C
$$< 2.8 \times 10^{-9}m$$

D
$$\geq 2.8 \times 10^{-9}m$$

Solution

From photoelectric effect, $$h\nu=W_0+eV$$

or $$\dfrac{hc}{\lambda}=W_0+eV$$

or $$\displaystyle \frac{(6.6\times 10^{-34})\times (3\times 10^8)}{500\times 10^{-9}}=2.28\times 1.6\times 10^{-19}+eV$$

or $$eV=0.31 \times 10^{-19}$$ or $$V=0.2$$

Here V should be $$V\leq 0.2$$

The de Broglie wavelength of electron , $$\lambda=\dfrac{12.27\times 10^{-10}}{V}$$

so, $$\lambda\geq 2.8\times 10^{-9} m$$
Question 5
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If the momentum of an electron is changed by $$P$$, then the de-Broglie wavelength associated with it changes by $$0.5%$$. The initial momentum of electron will be :

A
$$400P$$

B
$$\cfrac{P}{200}$$

C
$$100P$$

D
$$200P$$

Solution

Initial momentum, $$P_i=\cfrac{h}{\lambda}$$
or
$$P_i-P=\dfrac{h}{\lambda+\dfrac{0.5}{100}\lambda}$$
or
$$P_i-P=\dfrac{P_i}{\dfrac{1005}{1000}}$$
or
$$P_i=\dfrac{1005P}{5}\approx 200P$$
Question 6
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The wavelength $$\lambda _{e}$$ of an electron and $$\lambda _{p}$$ of a photon of same energy E are related by :

A
$$\lambda _{p}\propto \lambda _{e}^{2}$$

B
$$\lambda _{p}\propto \lambda _{e}$$

C
$$\lambda _{p}\propto \sqrt{\lambda _{e}}$$

D
$$\displaystyle \lambda _{p}\propto \dfrac{1}{\sqrt{\lambda _{e}}}$$

Solution

Wavelength of electron, $$\displaystyle \lambda _{e}=\dfrac{h}{\sqrt{2mE}}$$

Wavelength of photon, $$\displaystyle \lambda _{p}=\dfrac{hc}{E}$$

$$\displaystyle \lambda _{e}^{2}=\dfrac{h^{2}}{2mE}$$

$$\displaystyle \lambda _{e}^{2}=\dfrac{h^{2}}{2m\dfrac{hc}{\lambda _{p}}}\\ \Rightarrow \lambda _{e}^{2}\propto
\lambda _{p}$$
Question 7
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An electron of mass $$m$$ and a photon have same energy $$E$$. The ratio of de-Broglie wavelength associated with them is:

A
$$\dfrac { 1 }{ c } { \left( \dfrac { E }{ 2m } \right) }^{ \dfrac { 1 }{ 2 } }$$

B
$${ \left( \dfrac { E }{ 2m } \right) }^{ \dfrac { 1 }{ 2 } }$$

C
$$c{ \left( 2mE \right) }^{ \dfrac { 1 }{ 2 } }$$

D
$$\dfrac { 1 }{ c } { \left( \dfrac { 2m }{ E } \right) }^{ \dfrac { 1 }{ 2 } }$$

($$c$$ being velocity of light)

Solution

For an electron, $$\lambda_e=\dfrac{h}{\sqrt{2mE}}$$

For photon, $$E=pc\Rightarrow \lambda_{Photon}=\dfrac{hc}{E}$$

$$\Rightarrow \dfrac{\lambda_e}{\lambda_{Photon}}=\dfrac{h}{\sqrt{2mE}}\times \dfrac{E}{hc}=\left (\dfrac{E}{2m} \right )^{1/2}\dfrac{1}{c}$$
Question 8
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An electron is accelerated from rest through a potential difference of $$V$$ volt. If the de Broglie wavelength of the electron is $$1.227 \times 10^{-2} nm$$, the potential difference is :

A
$$10^2V$$

B
$$10^3V$$

C
$$10^4V$$

D
$$10V$$

Solution

The de-broglie wavelength of an electron is given as:
$$\lambda=\dfrac{1.227}{\sqrt{V}}\ nm$$

Substitute the wavelength in the above expression:
$$V=\left(\dfrac{1.227}{1.227\times10^{-2}}\right)^2$$

$$V=10^4\ V$$
Question 9
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If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is :

A
25

B
75

C
60

D
50

Solution
Question 10
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Which of the following figures represent the variation of particle momentum and the associated de-Broglie wavelength?

A

B

C

D

Solution

HINT: ACCORDING TO DE BROGLIE THE WAVELENGTH OF AN OBJECT IS ASSOCIATED WITH THE MOMENTUM AND MASS OF THE BODY

STEP 1: Write the formula of momentum.

$$\mathrm{p}=\dfrac{\mathrm{h}}{\lambda}$$

As $$\lambda$$ increases, $$p$$ decreases.

Step 2: Find the correct graph.

Option(A) is wrong since the slope is a straight line but according to the above relation, the graph can not be a straight line.
Differentiating the momentum.

$$\dfrac{d p}{d x}=-\dfrac{h}{\lambda^{2}}$$
as the value is negative therefore the graph should be decreasing.
Therefore, option (D) is the correct answer.