Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

Find the wavelength of 100 $$eV$$ electron

A
$$1.227 \displaystyle A^{\circ}$$

B
$$1.72 \displaystyle A^{\circ}$$

C
$$1.24\displaystyle nm $$

D
$$12.4 \displaystyle nm $$

Solution

We have

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

$$\implies \lambda = \dfrac {6.625 \times 10^{-34}}{\sqrt {2 \times 9.1 \times 10^{-31} \times 100 \times 1.6 \times 10^{-19}}} = 1.227 A^o$$

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

A surface has work function 3.3 $$eV$$ . Which of the following will cause emission?

A
100 $$W$$ incandascent lamp

B
40 $$W$$ flouroscent lamp

C
20 $$W$$ sodium lamp

D
20 $$W$$ Hg lamp

Solution

Since $$\lambda_{min} =400$$ is the minimum wavelength of visible region we will use it in the expression

$$E = \dfrac {1240}{\lambda} eV$$

$$\implies E = \dfrac{1242}{400} =3.1 eV.$$

$$\therefore$$ In order for emission we require a source of uv light, since visible light can not cause emission. In the given options, only Hg lamp is capable of generating uv light.
So, the answer is option (D).
Question 3
1 / -0

If the $$KE$$ of a free electron doubles then its de-Broglie wavelength changes by a factor

A
$$\displaystyle\dfrac{1}{2}$$

B
$$\displaystyle\dfrac{1}{\sqrt 2}$$

C
$$\displaystyle2$$

D
$$\displaystyle\sqrt 2$$

Solution

$$\displaystyle \lambda =\dfrac{h}{p}=\dfrac{h}{\sqrt{2\left ( KE \right )m}}$$

Thus when KE is doubled, the wavelength is changed by a factor of $$\dfrac{1}{\sqrt 2}$$
Question 4
1 / -0

The photoelectrons emitted from the surface of sodium metal are

A
Of speeds from zero to a certain maximum

B
Of same De-Broglie wavelength

C
Of same kinetic energy

D
Of same frequency

Solution

The photoelectric effect related with the incident light frequency $$(\nu)$$ by the following equation:
$$E_k=eV_s=h\nu-\phi$$,
where $$\phi$$ is the work function of the material and $$E_k$$ is the kinetic energy of the photo electron. So the photoelectrons emitted from the surface of sodium metal are of speeds from zero to a certain maximum depending on the incident photon energy.
So, the answer is option (A).
Question 5
1 / -0

If a proton and an electron have the same de Broglie wavelength, the ratio of the velocity of proton to the velocity of electron will be nearly

A
$$1$$

B
$$1840$$

C
$$\displaystyle\dfrac{1}{1840}$$

D
$$44$$

Solution

De Broglie wavelength is given by:

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

If two bodies (electron and proton) have the same De Broglie wavelength, they have the same momentum.
i.e.
$$p_{1} = p_{2}$$
$$m_{e}v_{e} = m_{p}v_{p}$$
i.e.
$$\dfrac{m_{e}}{m_{p}} = \dfrac{v_{p}}{v_{e}} = \dfrac{1}{1840}$$

So, the answer is option (C).
Question 6
1 / -0

The value of stopping potential for $$\lambda _2$$ in the following diagram is

A
$$v_1$$

B
-3V

C
$$V_2$$

D
-1V

Solution

From the above graph,
The value of stopping potential for the $$\lambda _2$$ in the following diagram is $$V_2$$.
Question 7
1 / -0

The necessary condition for photoelectric emission is :

A
$$\displaystyle h\nu < h\nu_{0}$$

B
$$\displaystyle h\nu > h\nu_{0}$$

C
$$\displaystyle E_{r} < h\nu_{0}$$

D
$$\displaystyle E_{k} > h\nu_{0}$$

Solution

Light with frequencies below a certain cutoff value, called the threshold frequency, would not eject photoelectrons from the metal surface no matter how bright the source is.
Hence $$h \nu > h \nu_o$$
Question 8
1 / -0

The hydrogen atom emits a photon of 656.3 nm line. Find the momentum of the photon associated with it.

A
$$ 10^{-27}$$ kg $$ms^{-1}$$

B
$$ 10^{-23}$$ kg $$ms^{-1}$$

C
$$ 10^{-25}$$ kg $$ms^{-1}$$

D
none of these

Solution

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

$$=\dfrac{6.625 \times10^{-34}}{656.3 \times 10^{-9}}$$

$$= 1.0 \times 10^{-27} kg ms^{-1}$$
So, the answer is option (A).
Question 9
1 / -0

A photon of energy $$h\nu$$ is absorbed by a free electron of a metal having work function $$\displaystyle \phi < h\nu.$$

A
The electron is sure to come out.

B
The electron is sure to come out with a kinetic energy $$\displaystyle hv-\phi.$$

C
Either the electron does not come out or it comes out with a kinetic energy $$\displaystyle hv-\phi.$$

D
If may come out with a kinetic energy less than $$\displaystyle hv-\phi.$$

Solution

For a photon with an energy larger than the work function ($$\phi$$), the excess energy is carried off by the electron as kinetic energy. Thus, the maximum kinetic energy is
$$ KE_{max} = h \nu - \phi$$
So, the answer is option (D).
Question 10
1 / -0

The energy that should be added to an electron to reduce its de broglie wavelength from 1 nm to 0.5 nm is

A
Four times the initial energy.

B
Equal to initial energy.

C
Twice the initial energy.

D
Thrice the initial energy.

Solution

$$\ \lambda =\dfrac{h}{p}=\dfrac{h}{\sqrt{2m(KE)}}$$

$$\therefore $$ New energy should be 4 times and hence energy to be added is thrice the initial energy.
So, the answer is option (D).

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Dual Nature of Radiation and Matter Test - 28

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Dual Nature of Radiation and Matter Test - 28

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Question 1

$$1.227 \displaystyle A^{\circ}$$

$$1.72 \displaystyle A^{\circ}$$

$$1.24\displaystyle nm $$

$$12.4 \displaystyle nm $$

Solution

Question 2

100 $$W$$ incandascent lamp

40 $$W$$ flouroscent lamp

20 $$W$$ sodium lamp

20 $$W$$ Hg lamp

Solution

Question 3

$$\displaystyle\dfrac{1}{2}$$

$$\displaystyle\dfrac{1}{\sqrt 2}$$

$$\displaystyle2$$

$$\displaystyle\sqrt 2$$

Solution

Question 4

Of speeds from zero to a certain maximum

Of same De-Broglie wavelength

Of same kinetic energy

Of same frequency

Solution

Question 5

$$1$$

$$1840$$

$$\displaystyle\dfrac{1}{1840}$$

$$44$$

Solution

Question 6

$$v_1$$

-3V

$$V_2$$

-1V

Solution

Question 7

$$\displaystyle h\nu < h\nu_{0}$$

$$\displaystyle h\nu > h\nu_{0}$$

$$\displaystyle E_{r} < h\nu_{0}$$

$$\displaystyle E_{k} > h\nu_{0}$$

Solution

Question 8

$$ 10^{-27}$$ kg $$ms^{-1}$$

$$ 10^{-23}$$ kg $$ms^{-1}$$

$$ 10^{-25}$$ kg $$ms^{-1}$$

none of these

Solution

Question 9

The electron is sure to come out.

The electron is sure to come out with a kinetic energy $$\displaystyle hv-\phi.$$

Either the electron does not come out or it comes out with a kinetic energy $$\displaystyle hv-\phi.$$

If may come out with a kinetic energy less than $$\displaystyle hv-\phi.$$

Solution

Question 10

Four times the initial energy.

Equal to initial energy.

Twice the initial energy.

Thrice the initial energy.

Solution

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