Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 37

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Question 1
1 / -0

When light of wave length 300 nm falls on a photoelectric emitter, photoelectrons are liberated. For another emitter, however, light of 600nm wavelength is sufficient for creating photoemission. What is the ratio of the work functions of the two emitters?

A
1:2

B
2:1

C
1:4

D
4:1

Solution

Work function   $$\phi = \dfrac{hc}{\lambda}$$
$$\implies$$ $$\dfrac{\phi_1}{\phi_2} = \dfrac{\lambda_2}{\lambda_1}$$
Given : $$\lambda_1 = 300 \ nm$$          $$\lambda_2 = 600 \ nm$$
$$\therefore$$   $$\dfrac{\phi_1}{\phi_2} = \dfrac{600}{300}=2$$
Question 2
1 / -0

The work function of aluminum is 4.2 eV. If two photons each of energy 3.5 eV strike an electron of aluminum, then emission of electron will be

A
depends on the density of the surface

B
not possible

C
possible

D
none of these

Solution

Emission of photo electrons from the metal surface takes place only when the energy of the incident photon is greater than the work function of the metal surface.
Since here, energy of incident photon is less than work function of metal surface. So emission of electron is not possible.
Question 3
1 / -0

An electron accelerated by a potential difference of V volt posses a de Broglie wave length . If the accelerating potential is increased by a factor of 4, the de-Broglie wavelength of the electron will be

A
remains unchanged

B
becomes double

C
becomes half

D
becomes 4 times

Solution

de-Broglie wavelength is given by $$\lambda = \dfrac{h}{p}$$
where, momentum of the particle $$p = \sqrt{2mK}$$
Kinetic energy of the particle $$K = eV$$
$$\implies \ \lambda = \dfrac{h}{\sqrt{2meV}}$$
We get $$\lambda\propto \dfrac{1}{\sqrt{V}}$$
Thus de-Broglie wavelength of the electron will become half if the accelerating potential is increased by a factor of $$4$$.
Question 4
1 / -0

Monochromatic light of wave length 667 nm is produced by a helium neon laser. The power emitted is 9mW. The number of photons arriving per second on the average at a target irradiated by this beam is

A
$$3\times{10}^{16}$$

B
$$3\times{10}^{19}$$

C
$$9\times{10}^{15}$$

D
$$9\times{10}^{17}$$

Solution

Wavelength of light $$\lambda = 667 \ nm$$
Energy of incident photon $$E = \dfrac{hc}{\lambda}$$
$$\implies \ E = \dfrac{6.626\times 10^{-34}\times 3\times 10^8}{667\times 10^{-9}} = 3\times 10^{-19} \ J$$
Power of light $$P = 9mW = 0.009 \ W$$
So, number of photons per second $$N \dfrac{P}{E} = \dfrac{0.009}{3\times 10^{-19}} = 3\times 10^{16}$$ photons/ second
Question 5
1 / -0

According to Einstein's interpretation of the photoelectric effect, the maximum K.E. of photoelectrons depends on ( h -W) where is the frequency of incident radiation and W is the work function. For three different metals graph plotted between maximum K/E. and the frequency of incident radiations. The three graphs obtained:

A
have different slopes but constant intercept on y axis

B
are all parallel to y axis but at different heights

C
same slope but different intercept on y axis

D
none of these

Solution

According to Einstein's equation,
$$ h\nu = W + {K.E}_{max} $$
$$ K.E._{max} = h\nu - W$$

Comparing it with equation of straight line with slope $$ m $$ and intercept $$ c $$,
$$ y = mx + c $$

We see that on taking $$ K.E_{max} $$ on $$ y-axis $$ and frequency of incident radiation on $$ x - axis$$ , we get a straight line with intercept $$ W = work function$$ and slope $$ h = \text{Planck's constant}$$

We see that while plotting the graph of $$ K.E. $$ vs $$ \nu$$ for 3 different metals , slope will be same as $$ h $$ is same for all 3 metals. But $$ W $$ is different for different metals . So intercept will be different for different metals.
Question 6
1 / -0

According to Einstein's photoelectric equation, the graph of K.E. of the photoelectron emitted from the metal versus the frequency of the incident radiation gives a straight line graph, whose slope

A
is same for all metals and independent of the intensity of the incident radiation.

B
depends on the nature of the metal.

C
depends on the intensity of the incident radiation.

D
depends on the nature of the metal and also on the intensity of incident radiation.

Solution

Kinetic energy of photo electron emitted $$K.E = h\nu -\phi$$
where $$\nu$$ is the frequency of incident radiation and $$\phi$$ is the work function of the metal.
Thus graph of $$K.E$$ Vs $$\nu$$ is a straight line with slope equal to $$h$$, having an intercept on negative y-axis which is equal to the work function $$(\phi)$$ of metal.
As $$h$$ is a constant, thus the slope is same for all metals and independent of the intensity of the incident radiation.
Question 7
1 / -0

If the kinetic energy of the moving particle is $$E$$, then the de Broglie wavelength is

A
$$\lambda = \dfrac {h}{\sqrt {2mE}}$$

B
$$\lambda = \dfrac {\sqrt {2mE}}{h}$$

C
$$\lambda = h \sqrt {2mE}$$

D
$$\lambda = \dfrac {h}{E\sqrt {2m}}$$

Solution

De Broglie wavelength of the particle, $$\lambda = \dfrac{h}{p}$$ where $$p$$ is the momentum of the particle

Kinetic energy of particle, $$E = \dfrac{1}{2}mv^2 = \dfrac{p^2}{2m}$$ $$(\because p = mv)$$

Thus, we get: $$p = \sqrt{2mE}$$

$$\implies$$ $$\lambda = \dfrac{h}{\sqrt{2mE}}$$
Question 8
1 / -0

The threshold frequency for a photo-sensitine metal is $$3.3\times { 10 }^{ 14 }Hz$$. If light of frequency $$8.2\times { 10 }^{ 14 }Hz$$ is incident on this metal, the cut-off voltage for the photo-electric emission is nearly

A
$$2 V$$

B
$$3 V$$

C
$$5 V$$

D
$$1 V$$

Solution

Here, $${ V }_{ 0 }=\dfrac { E-V }{ e } =\dfrac { h\left( v-{ v }_{ 0 } \right) }{ e }$$
$$ =\dfrac { 6.62\times { 10 }^{ -34 }\left( 8.2\times { 10 }^{ 14 }-3.3\times { 10 }^{ 14 } \right) }{ 1.6\times { 10 }^{ -19 } }$$
$$ =\dfrac { 6.62\times { 10 }^{ -34 } }{ 1.6 } \times 4.9\times { 10 }^{ 33 }$$
$$=\dfrac { 6.62\times 4.9\times { 10 }^{ -1 } }{ 1.6 }$$
$$ { V }_{ 0 }=2$$ volt
Question 9
1 / -0

The kinetic energy of an electron get tripled then the de-Broglie wavelength associated with electron changes by a factor of

A
$$\dfrac { 1 }{ 3 } $$

B
$$\sqrt { 3 } $$

C
$$\dfrac { 1 }{ \sqrt { 3 } } $$

D
$$3$$

Solution

de-Broglie wavelength of an electron $$\lambda =\dfrac { h }{ \sqrt { 2mK } }$$
Or $$ \lambda \propto \dfrac { 1 }{ \sqrt { K } }$$
where $$K$$ is the kinetic energy of the electron and $$m$$ is the mass.
$$ \therefore \dfrac { { \lambda }^{ \prime } }{ \lambda } =\dfrac { 1 }{ \sqrt { 3K } } .\dfrac { \sqrt { K } }{ 1 } =\dfrac { 1 }{ \sqrt { 3 } }$$
Or $$ { \lambda }^{ \prime }=\dfrac { \lambda }{ \sqrt { 3 } }$$
i.e., de-Broglie wavelength will decrease by a factor of $$ \dfrac { 1 }{ \sqrt { 3 } } $$.
Question 10
1 / -0

Photons of energy $$7 e V$$ are incident on two metals A and B with work functions $$6 eV$$ and $$3 eV$$ respectively. The minimum de Broglie wavelengths of the emitted photoelectrons with maximum energies are $$\lambda_A$$ and $$\lambda_B$$ , respectively where $$\lambda_A / \lambda_B $$ is nearly :

A
$$0.5$$

B
$$1.4$$

C
$$4.0$$

D
$$2.0$$

Solution

Kinetic energy of photo electrons $$K = E_{photons} - \phi$$
where $$\phi$$ is the work function of the metal and $$E_{photon}$$ is the energy of the photon.
For metal A : $$K_A = 7-6 =1$$ eV
For metal B : $$K_B = 7-3 =4$$ eV
de Broglie wavelength $$\lambda = \dfrac{h}{\sqrt{2mK}}$$
$$\implies$$ $$\dfrac{\lambda_A}{\lambda_B} = \sqrt{\dfrac{K_B}{K_A}}$$
Or $$\dfrac{\lambda_A}{\lambda_B} = \sqrt{\dfrac{4}{1}} = 2.0$$

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Dual Nature of Radiation and Matter Test - 37

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Dual Nature of Radiation and Matter Test - 37

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Question 1

1:2

2:1

1:4

4:1

Solution

Question 2

depends on the density of the surface

not possible

possible

none of these

Solution

Question 3

remains unchanged

becomes double

becomes half

becomes 4 times

Solution

Question 4

$$3\times{10}^{16}$$

$$3\times{10}^{19}$$

$$9\times{10}^{15}$$

$$9\times{10}^{17}$$

Solution

Question 5

have different slopes but constant intercept on y axis

are all parallel to y axis but at different heights

same slope but different intercept on y axis

none of these

Solution

Question 6

is same for all metals and independent of the intensity of the incident radiation.

depends on the nature of the metal.

depends on the intensity of the incident radiation.

depends on the nature of the metal and also on the intensity of incident radiation.

Solution

Question 7

$$\lambda = \dfrac {h}{\sqrt {2mE}}$$

$$\lambda = \dfrac {\sqrt {2mE}}{h}$$

$$\lambda = h \sqrt {2mE}$$

$$\lambda = \dfrac {h}{E\sqrt {2m}}$$

Solution

Question 8

$$2 V$$

$$3 V$$

$$5 V$$

$$1 V$$

Solution

Question 9

$$\dfrac { 1 }{ 3 } $$

$$\sqrt { 3 } $$

$$\dfrac { 1 }{ \sqrt { 3 } } $$

$$3$$

Solution

Question 10

$$0.5$$

$$1.4$$

$$4.0$$

$$2.0$$

Solution

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