Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

An electron and a proton are accelerated through the same potential difference. The ratio of their de Broglie wavelengths ($$\dfrac{\lambda_{e}}{\lambda_{p}}$$ ) is

A
1

B
$$\dfrac{m_{e}}{m_{p}}$$

C
$$\dfrac{m_{p}}{m_{e}}$$

D
$$\sqrt{\dfrac{m_{p}}{m_{e}}}$$

Solution

$$\lambda _{e}=\dfrac{h}{\sqrt{2m_{e}eV}}$$

$$\lambda _{p}=\dfrac{h}{\sqrt{2m_{p}eV}}$$

$$\dfrac{\lambda _{e}}{\lambda _{p}}=\sqrt{\dfrac{m_{p}}{m_{_{e}}}}$$

So, the answer is option (D).
Question 2
1 / -0

The de-Broglie wavelength associated with an electron accelearated to potential difference of V volt is:

A
$$V\times 10^{-10}m$$

B
$$\sqrt{\dfrac{150}{V}}A^{0}$$

C
$$\dfrac{150}{V}A^{0}$$

D
$$150V\times 10^{-8}m$$

Solution

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

$$p=\sqrt{2meV}$$

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

$$=\dfrac{h}{\sqrt{2meV}}$$

$$=\sqrt{\dfrac{150}{V}} A^o$$
So, the answer is option (B).
Question 3
1 / -0

The de Broglie wavelength of a neutron when its kinetic energy is K is $$\lambda$$ .What will be its wavelength when its kinetic energy is 4K ?

A
$$\dfrac{\lambda}{4}$$

B
$$\dfrac{\lambda}{2}$$

C
2$$\lambda$$

D
4$$\lambda$$

Solution

if KE has become 4 times, then speed must have doubled.

and $$\lambda = \dfrac{h}{mV}$$

i.e. $$\lambda \alpha \dfrac{1}{V}$$

$${\lambda}_1 V_1 = {\lambda}_2 V_2$$
$$\lambda V = \lambda (2V)$$

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

So, the answer is option (B).
Question 4
1 / -0

The de Broglie wavelength associated with an electron of velocity 0.3c and rest mass $$9.1 \times 10^{-31}$$kg is

A
$$7.68\times 10^{-10}m$$

B
$$7.68\times 10^{-12}m$$

C
$$5.7\times 10^{-12}m$$

D
$$9.1\times 10^{-12}m$$

Solution

The de-broglie wavelength according to formula $$=\dfrac{h}{m_ov}\sqrt{(1-\dfrac{v^2}{c^2})}$$

$$\Rightarrow \lambda =\dfrac{6.63\times10^{-34}}{9.1\times10^{-31}\times0.3\times3\times10^{8}}\sqrt{1-(\dfrac{0.3c}{c})^2}$$

$$\Rightarrow \lambda =7.68\times10^{-12}m.$$
So, the answer is option (B).
Question 5
1 / -0

A proton when accelerated through a potential difference of V volt has wavelength $$\lambda$$ associated with it .An electron to have the same $$\lambda$$ must be accelerated through a p.d of

A
$$\dfrac{V}{8}$$ volt

B
4V volt

C
2V volt

D
1838V volt

Solution

The energy gained by the proton will be
$$E = q_oV = \dfrac{p^2}{2m_p}$$ ...... ( 1 )
where $$q_o$$ is the charge on the proton, V is the p.d applied, p is the momentum of the proton and $$m_p$$ is the mass of the proton.
The momentum p of the proton is given by de Broglie's equation as
$$p = \dfrac{h}{\lambda}$$
Therefore, for an electron to have the same $$\lambda$$, it must have the same momentum.
$$\therefore$$ $$E_e = \dfrac{p^2}{2m_e}$$
Using p from equation 1,
$$\implies$$ $$E_e = \dfrac{m_pq_oV}{m_e}$$
$$\implies$$ $$E_e = q_oV(\dfrac{m_p}{m_e}) = q_o(1838V)$$
Therefore, an electron must be accelerated through a p.d of 1838V volt in order to achieve a wavelength $$\lambda$$.
Question 6
1 / -0

The de Broglie wavelength of an electron which falls through a p.d. of 10,000V is

A
1.227 $$A^{0}$$

B
12.27 $$A^{0}$$

C
0.1227 $$A^{0}$$

D
2.454 $$A^{0}$$

Solution

$$\lambda = {\dfrac{h}{\sqrt{2MqV}}}$$
$$={\dfrac{6.62\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 10^4}}}$$
$$= 0.1227 A^o$$
So, the answer is option (C).
Question 7
1 / -0

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

A
$$3\times10^{19}$$

B
$$9\times10^{17}$$

C
$$3\times10^{16}$$

D
$$9\times10^{15}$$

Solution

$$No. of photons =\frac{E}{(hc/\lambda)}$$

    $$=\frac{9\times10^{-3}\times6.67\times10^{-7}}{6.6\times10^{-34}\times3\times10^{8}}$$

   $$=3\times10^{16}$$
Question 8
1 / -0

The velocity of electron of H-atom in its ground state is $$2.2\times 10^{-6} m/s$$. The de-Broglie wavelength of this electron would be:

A
$$0.33 \ nm$$

B
$$23.3 \ nm$$

C
$$45.6\ nm$$

D
$$100 \ nm$$

Solution

Given:
$$v= 2.2\times 10^{-6} m/s, m_e=9.1\times 10^{31}, h= 6.61 \times10^{-34}$$

$$\displaystyle \lambda=\frac {h}{mv}=\frac {6.626\times 10^{-34}}{9.1\times 10^{-31}\times 2.2\times 10^{-6}}=0.33\times 10^{-9}=0.33 nm$$

Hence, the correct option is A.
Question 9
1 / -0

The de-Broglie wavelength of an electron traveling with speed equal to 1% of the speed of light:

A
400 pm

B
120 pm

C
242 pm

D
375 pm

Solution

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

$$v=\dfrac{1}{100}\times c=\dfrac{1}{100}\times 3\times10^8=3\times10^6 m/s.$$

$$\lambda=\dfrac{6.626\times10^{-34}}{9.1\times10^{-31}\times3\times10^6}=242 pm$$

Hence, option C is correct.
Question 10
1 / -0

Tungsten has work function 4.8 $$eV$$ . We wish to use tungsten as photo-cathode with a 600 $$nm$$ wavelength. What shall we do?

A
Coat tungsten with Cesium

B
Oxide coat tungsten

C
$$Cu_{2}$$ $$O_{2}$$ be coated on tungsten

D
None of these

Solution

A photocathode usually consists of alkali metals with very low work function. The coating releases electrons much more readily than the underlying metal, allowing it to detect the low-energy photons in infrared radiation. To lowering the work function of the tungsten Oxide coating is used.
So, the answer is option (B).

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Dual Nature of Radiation and Matter Test - 27

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Dual Nature of Radiation and Matter Test - 27

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Question 1

1

$$\dfrac{m_{e}}{m_{p}}$$

$$\dfrac{m_{p}}{m_{e}}$$

$$\sqrt{\dfrac{m_{p}}{m_{e}}}$$

Solution

Question 2

$$V\times 10^{-10}m$$

$$\sqrt{\dfrac{150}{V}}A^{0}$$

$$\dfrac{150}{V}A^{0}$$

$$150V\times 10^{-8}m$$

Solution

Question 3

$$\dfrac{\lambda}{4}$$

$$\dfrac{\lambda}{2}$$

2$$\lambda$$

4$$\lambda$$

Solution

Question 4

$$7.68\times 10^{-10}m$$

$$7.68\times 10^{-12}m$$

$$5.7\times 10^{-12}m$$

$$9.1\times 10^{-12}m$$

Solution

Question 5

$$\dfrac{V}{8}$$ volt

4V volt

2V volt

1838V volt

Solution

Question 6

1.227 $$A^{0}$$

12.27 $$A^{0}$$

0.1227 $$A^{0}$$

2.454 $$A^{0}$$

Solution

Question 7

$$3\times10^{19}$$

$$9\times10^{17}$$

$$3\times10^{16}$$

$$9\times10^{15}$$

Solution

Question 8

$$0.33 \ nm$$

$$23.3 \ nm$$

$$45.6\ nm$$

$$100 \ nm$$

Solution

Question 9

400 pm

120 pm

242 pm

375 pm

Solution

Question 10

Coat tungsten with Cesium

Oxide coat tungsten

$$Cu_{2}$$ $$O_{2}$$ be coated on tungsten

None of these

Solution

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