Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 51

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Weekly Quiz Competition

Question 1
1 / -0

A photoelectric cell is lightened by a light source, situated at a distance d from the cell. If distance becomes $$d/2$$ then number of electrons emitted per sec will be:-

A
Remains same

B
Four times

C
Two times

D
One fourth

Solution

Intensity is directly proportional to number of photons $$incident$$ on the surface.
$$Intensity \propto \dfrac{1}{distance^2}$$
so on making distance $$1/2 $$ we will get intensity $$(\dfrac{1}{1/2})^2=4$$ $$times$$
One photon contribute in one electrons so the number of electrons will also become $$4$$ times.
Question 2
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In stopping potential ($$V$$) photo current ($$I$$) graph, if $${V}_{2}> {V}_{1}$$ then compare the wavelengths of incident radiations

A
$${\lambda}_{1}=\sqrt{{\lambda}_{2}}$$

B
$${\lambda}_{1}< {\lambda}_{2}$$

C
$${\lambda}_{1}= {\lambda}_{2}$$

D
$${\lambda}_{1}> {\lambda}_{2}$$

Solution
Question 3
1 / -0

The de-Broglie wavelength of a proton accelerated by $$400\ V$$ is

A
$$0.005\ \mathring {A}$$

B
$$1.0528\ \mathring {A}$$

C
$$0.0568\ \mathring {A}$$

D
$$0.0143\ \mathring {A}$$

Solution

$$0.043\dot{A}$$

formulae,

$$qv=\dfrac{1}{2}mv^2$$

$$\Rightarrow v=\sqrt{\dfrac{2qV}{m}}$$

$$\Rightarrow v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 400}{1.67\times 10^{-27}}}=2.77\times 10^{5}m/s$$

$$\lambda =\dfrac{h}{mv}$$

$$=\dfrac{6.63 \times 10^{-34}}{(1.67\times 10^{-27})(2.77\times 10^{5})}$$

$$\therefore \lambda =1.43\times 10^{-12}m$$

$$\Rightarrow \lambda =0.0143\dot{A}$$
Question 4
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Find the number of electrons emitted per second by a $$24 W$$ source of monochromatic light of wavelength $$6600 A$$ assuming $$3%$$ efficiency for photoelectric effect (take $$h = 6.6 \times 10_34 \, Js$$)

A
$$48 \times 10^{19}$$

B
$$24 \times 10^{17}$$

C
All of these

D
None of these

Solution
Question 5
1 / -0

A proton and an electron are accelerated by same potential difference starting from rest have de- Brogile wavelength $$\lambda _ { p }$$ and $$\lambda _ { e ^ { * } }$$

A
$$\lambda _ { e } = \lambda _ { p }$$

B
$$\lambda _ { 0 } < \lambda _ { p }$$

C
$$\lambda _ { e } > \lambda _ { p }$$

D
none of these

Solution

The wavelength of the photon is given as:
$$\begin{array}{l} \lambda =\dfrac { h }{ P } \\ =\dfrac { h }{ { \sqrt { 2mk } } } \\ \lambda \propto \dfrac { 1 }{ { \sqrt { m } } } \\ \therefore { \lambda _{ p } }<{ \lambda _{ e } } \\ Hence, \ option\, \, C\, \, is\, \, correct\, \, answer. \end{array}$$
Question 6
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For $$\gamma$$ photon emission, Which is in correct ?

A
$$ \ \gamma \ emission \ may\ follow \propto - emission $$

B
$$\gamma\ emission\ may \ follow \beta - emission$$

C
$$\gamma \ emission \ must \ follow \propto - emission $$ and no \$$beta-emission $$

D
$$speed \ of \gamma\ rays\ is\ equal \ to the\ speed \ of \ light $$

Solution
Question 7
1 / -0

The phenomenon of photo emission depends on-

A
Only wave length of incident light

B
Only work function of surface

C
Only nature of surface

D
All of the above

Solution
Question 8
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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 held on, and elastic. The ratio of the de-Broglie wavelength $$\lambda_{A}$$ to $$\lambda_{B}$$ after the collision is

A
$$\dfrac {\lambda_{A}}{\lambda_{B}} = 2$$

B
$$\dfrac {\lambda_{A}}{\lambda_{B}} = \dfrac {2}{3}$$

C
$$\dfrac {\lambda_{A}}{\lambda_{B}} = \dfrac {1}{2}$$

D
$$\dfrac {\lambda_{A}}{\lambda_{B}} = \dfrac {1}{3}$$

Solution

By conservation of linear momentum $$mV={ mV }_{ 1 }+\dfrac { m }{ 2 } { V }_{ 2 }$$
$$2V=2{ V }_{ 1 }+{ V }_{ 2 }\quad \longrightarrow \left( 1 \right) $$ $${ \lambda }_{ 1 }=\dfrac { h }{ { p }_{ 1 } } $$ ; $${ \lambda }_{ 2 }=\dfrac { h }{ { O }_{ 2 } } $$
By law of collision
$$e=\dfrac { { V }_{ 2 }-{ V }_{ 1 } }{ { U }_{ 1 }-{ U }_{ 2 } } $$ $$\dfrac { { \lambda }_{ 1 } }{ { \lambda }_{ 2 } } =\dfrac { 2 }{ 1 } $$
$$u={ V }_{ 2 }-{ V }_{ 1 }\quad \longrightarrow \left( 2 \right) $$ $$\boxed { { \lambda }_{ 1 }:{ \lambda }_{ 2 }=2:1 } $$
equ $$(1)$$ and $$(2)$$
$${ V }_{ 1 }=\dfrac { V }{ 3 } \quad ;\quad { V }_{ 2 }=\dfrac { 4V }{ 3 } $$
Question 9
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A heavy nucleus at rest breaks into two fragments which fly off with velocities in the ratio $$8:1$$. the ratio of de-broglie wavelengths of fragments are

A
$$1:2$$

B
$$1:8$$

C
$$4:1$$

D
None of these

Solution

Momentum conversation. Given,
$${ m }_{ 1 }{ V }_{ 1 }={ m }_{ 2 }{ V }_{ 2 }$$
$$\dfrac { { V }_{ 1 } }{ { V }_{ 2 } } =\dfrac { 8 }{ 1 } =\dfrac { { m }_{ 2 } }{ { m }_{ 1 } } $$
So, $$\dfrac { { m }_{ 1 } }{ { m }_{ 2 } } =\dfrac { 1 }{ 8 } $$
but when particle break then density $$=$$ constant
So, $$V\alpha m$$
$$\dfrac { { V }_{ 1 } }{ { V }_{ 2 } } =\dfrac { { m }_{ 1 } }{ { m }_{ 2 } } $$
$$\dfrac { { a }_{ 1 }^{ 3 } }{ { a }_{ 2 }^{ 3 } } =\dfrac { 1 }{ 8 } $$
$$\dfrac { { a }_{ 1 } }{ { a }_{ 2 } } ={ \left( \dfrac { 1 }{ 8 } \right) }^{ 1/3 }\Rightarrow \boxed { \dfrac { { a }_{ 1 } }{ { a }_{ 2 } } =\dfrac { 1 }{ 2 } } $$
$$\Rightarrow { a }_{ 1 }:{ a }_{ 2 }=1:2$$
Question 10
1 / -0

Graph is plotted between maximum kinetic energy of electron with frequency of incident photon in Photo electric effect. The slope of curve will be :

A
Charge of electron

B
Work function of metal

C
Planck's constant

D
Ratio of planck constant and charge of electron

Solution

From the $$E_k-V$$ graph,
The slope of the curve is Planck's constant.
From the Einstein photoelectric equation,
$$E_k=hV-\phi$$. . . . .(1)
where, $$\phi=$$work function.
$$V=$$ frequency
From the equation (1)
$$m=h$$ (equation of straight line, $$Y=mx+c$$)
The slope of the curve is $$h$$.
The correct option is C.