Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 52

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Weekly Quiz Competition

Question 1
1 / -0

$${ 10 }^{ -3 }W$$ of 5000 A light is directed on a photoelectric cell. If the current in the cell is 0.16 $$\mu A$$, the percentage of incident photons which produce photo electrons, is

A
0.4%

B
0.04%

C
20%

D
10%

Solution

Current is $$0.16\times 10^{-6} Amp $$ it means $$0.16 \times 10^{-6} Coulomb$$ charge is flowing $$\text{per second}$$
So $$n=\dfrac{0.16\times 10^{-6}C}{1.6\times 10^{-19} C}=10^{12} \text{ electrons are generated per second}$$
Now we notice that $$\text{one photon has energy E}$$, $$E=\dfrac{hc}{\lambda}=\dfrac{6.62\times 10^{-34}Js \times 3\times 10^8 m/s}{5000\times 10^{-10}m}=3.972\times 10^{-19} Joule$$
So number of photon in $$10^{-3}W$$ will be $$N=\dfrac{10^{-3}}{3.972\times 10^{-19}}=0.25\times 10^{16}$$ this is number of $$\text{photons incident per second}$$
So required percentage is $$\dfrac{n}{N}\times 100=\dfrac{10^{14}}{0.25\times 10^{16}}=.04$$%

Option B is correct.
Question 2
1 / -0

The energy of a neutron in eV whose de-Broglie wavelength is $$1\mathop {\text{A}}\limits^{\text{0}} $$

A
$$1.67 \times {10^{ - 27}}eV$$

B
$$8.13 \times {10^{ - 2}}eV$$

C
$$6.62 \times {10^{ - 22}}eV$$

D
$$3.23 \times {10^{ - 2}}eV$$

Solution

Neutron $$ m = 1.67\times 10^{-27}$$ kg
$$ l = \frac{h}{p} $$ $$\Rightarrow P = \frac{h}{\lambda } $$ $$ = \frac{66\times 10^{-34}}{1\times 10^{-10}}$$ kg m/s
$$ \Rightarrow P = 6.6\times 10^{-24}$$ kg m/s
energy $$ E = \frac{P^{2}}{2m}= \frac{6.6\times 6.6\times 10^{-48}}{2\times 1.67\times 10^{-27}} $$ joule
$$ = \frac{(6.6)^{2}}{3.34}\times 10^{-21}\times \frac{1}{1.6\times 10^{-19}}$$ ev
E = 0.8 ev
Question 3
1 / -0

If a strong diffraction peak is observed when electron are incident at an angle. 'i' from the normal to the crystal planes with distance 'd' wavelength them (see figure) de Broglie wavelength $$ \lambda_{dB} $$ of electrons can be calculate by the relationship (n is an integer)

A
$$ d sin = n \lambda_{dB} $$

B
$$2 d cos i= n \lambda_{dB} $$

C
$$ d sin = i \lambda_{dB} $$

D
$$ d cos = i \lambda_{dB} $$

Solution

Bragg's relation $$n\lambda =2d\sin\theta $$ for having an in density maximum for diffraction pattern. But as the angle of incidence is given their $$n\lambda =2dcosi$$ is the formula for finding a peat.
Question 4
1 / -0

Two large parallel plates are connected with the terminal of $$100V$$ power supply. These plates have a fine hole at the centre. An electron having energy $$200eV$$ is so directed that it passes through the holes. When it comes out it's de-Broglie wavelength is

A
$$1.22 A ^ { \circ }$$

B
$$1.75 A ^ { \circ }$$

C
$$2 A ^ { \circ }$$

D
none of these

Solution

Energy of the electron, when it comes out from the second plate $$=200\ eV-100\ eV=100\ eV$$
Hence acceleration potential difference $$=100\ V$$
$$\lambda_{Electron}=\dfrac{12.27}{\sqrt{V}}=\dfrac{12.27}{\sqrt{100}}=1.23\overset{o}{A} $$
Question 5
1 / -0

The figure shows the path of white light's rays which leave in phase from two small source S$$_{1}$$ and S$$_{2}$$ and travel to a point X on a screen .The path difference is S$$_{2}$$X -- S$$_{1}$$X = 10 $$\times 10^{-7}$$m.What wavelength of light give complete destructive interference at X?

A
4.0$$\times 10^{-7}$$ m

B
6.6$$\times 10^{-7}$$ m

C
4.4$$\times 10^{-7}$$ m

D
5.8$$\times 10^{-7}$$ m
Question 6
1 / -0

Two particles A and B have de-Broglie's wavelengths $$30 \ \mathring A$$ and $$20 \ \mathring A$$, combined to form a particle C. Momentum is conserved in this process. The possible de-Broglie's wavelength of C is :
(the motion is one dimensional)

A
$$12 \ \mathring A$$

B
$$20 \ \mathring A$$

C
$$10 \ \mathring A$$

D
$$22 \ \mathring A$$

Solution

Given,

$$\lambda _A=30\times 10^{-10}m$$

$$\lambda _B=20\times 10^{-10}m$$

Momentum is conserved, so by applying the conservation of momentum,

$$P_C=P_A+P_B$$

$$\dfrac{h}{\lambda _C}=\dfrac{h}{\lambda _A}+\dfrac{h}{\Lambda _B}$$

$$\dfrac{1}{\lambda _C}=\dfrac{1}{30\times 10^{-10}}+\dfrac{1}{20\times 10^{-10}}$$

$$\lambda _C=12\times 10^{-10}m$$
Question 7
1 / -0

The energy that should be added to an electron to reduce its De broglie wavelength from $$10^{-10}$$ m to $$0.5 \times 10^{-10}$$ m will be :

A
Four times the initial energy

B
Equal to initial energy

C
Twice the initial energy

D
Thrice the initial energy

Solution

As we know,
$$\lambda =\dfrac {h}{\sqrt{2mE}}\Rightarrow \lambda \propto \dfrac {1}{\sqrt E}\Rightarrow \dfrac {\lambda_1}{\lambda_2}=\sqrt{\dfrac {E_2}{E_1}}$$
$$\Rightarrow \dfrac{10^{-10}}{0.5 \times 10^{-10}}=\sqrt{\dfrac {E_2}{E_1}}\Rightarrow E_2=4E_1$$
Hence added energy $$=E_2-E_1=3E_1$$
Question 8
1 / -0

A photocell is illuminated by a small bring source placed $$1 m$$ away. When the same source of light is placed $$1/2\ m$$ away, the number of electrons emitted by photocathode would :

A
Decrease by a factor of 4

B
Increase by a factor of 4

C
Decrease by a factor of 2

D
Increase by a factor of 2

Solution

Since Intensity $$\left( I \right) \alpha \dfrac { 1 }{ { r }^{ 2 } } $$

When distance become half, intensity becomes $$4$$ times. Hence number of electrons increases by factor of $$4$$. The number of electrons emitted of photo cathode would increase by a factor of $$4$$.
Question 9
1 / -0

If the de Broglie wavelengths associated with a proton and an $$\alpha$$-particle are equal then the ratio of velocities of the proton and the $$\alpha$$-particle will be:

A
$$4:1$$

B
$$2:1$$

C
$$1:2$$

D
$$1:4$$

Solution

De - Broglie wavelength
$$\lambda =\dfrac { h }{ p } =\dfrac { h }{ mv } =\dfrac { h }{ \sqrt { 2mE } } $$
wherein $$-h=$$ plank's constant
$$m=$$ mass of particle, $$v=$$ speed of particle
$$E=$$ Kinetic energy of particle
$$\lambda =\dfrac { h }{ mv } $$ $$\because$$ $${ \lambda }_{ \alpha }={ \lambda }_{ p }$$ $$\therefore$$ $${ p }_{ a }={ p }_{ b }$$
$$\therefore$$ $$m\alpha { v }_{ \alpha }={ m }_{ p }{ v }_{ p }$$
$$\therefore$$ $$\dfrac { { v }_{ p } }{ { v }_{ \alpha } } =\dfrac { { m }_{ a } }{ { m }_{ p } } =4:1$$
Question 10
1 / -0

If the particles listed below all have the same kinetic energy, which one would posses the shortest de-Broglie wavelength.

A
Electron

B
$$\alpha -$$particle

C
Proton

D
All of the above

Solution

As per the formula,
$$\lambda=\dfrac{h}{p}$$
Since $$\lambda=\dfrac{h}{mv}$$
thus, $$\lambda\propto\dfrac{1}{\sqrt{m}}$$
Wavelength and mass has inverse dependence.
Since, mass of electron is the least out of all the three particles $$\therefore$$, electron has the largest de-Broglie wavelength.

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Dual Nature of Radiation and Matter Test - 52

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Dual Nature of Radiation and Matter Test - 52

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Question 1

0.4%

0.04%

20%

10%

Solution

Question 2

$$1.67 \times {10^{ - 27}}eV$$

$$8.13 \times {10^{ - 2}}eV$$

$$6.62 \times {10^{ - 22}}eV$$

$$3.23 \times {10^{ - 2}}eV$$

Solution

Question 3

$$ d sin = n \lambda_{dB} $$

$$2 d cos i= n \lambda_{dB} $$

$$ d sin = i \lambda_{dB} $$

$$ d cos = i \lambda_{dB} $$

Solution

Question 4

$$1.22 A ^ { \circ }$$

$$1.75 A ^ { \circ }$$

$$2 A ^ { \circ }$$

none of these

Solution

Question 5

4.0$$\times 10^{-7}$$ m

6.6$$\times 10^{-7}$$ m

4.4$$\times 10^{-7}$$ m

5.8$$\times 10^{-7}$$ m

Question 6

$$12 \ \mathring A$$

$$20 \ \mathring A$$

$$10 \ \mathring A$$

$$22 \ \mathring A$$

Solution

Question 7

Four times the initial energy

Equal to initial energy

Twice the initial energy

Thrice the initial energy

Solution

Question 8

Decrease by a factor of 4

Increase by a factor of 4

Decrease by a factor of 2

Increase by a factor of 2

Solution

Question 9

$$4:1$$

$$2:1$$

$$1:2$$

$$1:4$$

Solution

Question 10

Electron

$$\alpha -$$particle

Proton

All of the above

Solution

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