Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

The wavelength associated with 1 $$MeV$$ proton is

A
28.6 $$pm$$

B
2.86 $$pm$$

C
2.86 $$fm$$

D
28.6 $$fm$$

Solution

The wavelength and the energy is related by:

$$\displaystyle \lambda =\dfrac{0.286}{\sqrt{10^{6}}}=28.6fm$$

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

The minimum energy required to dissociate $$Ag$$ $$Br$$ bond in 0.6 $$eV$$. A photographic flim is coated with a sliver bromide layer. Find the maximum wavelength whose signature can be recorded on the film

A
207 $$nm$$

B
702 $$nm$$

C
207 $$A^{\circ}$$

D
2070 $$nm$$

Solution

We have

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

$$\implies \lambda = \dfrac {1240}{E} = \dfrac {1240}{0.6} = 2070 nm$$

So, the answer is option (D).
Question 3
1 / -0

Find the wavelength of 10 $$MeV \ \alpha-$$particles

A
3 $$A^{\circ}$$

B
3 $$pm$$

C
3 $$fm$$

D
30 $$fm$$

Solution

The wavelength and the energy is related by:

$$\displaystyle \lambda (A^{\circ})=\frac{0.286}{\sqrt{10^{7}}}=0.3\times10^{-3}A^{\circ}=3\times10^{-15}m=3fm$$

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

When photons of energy $$h\nu$$ are incident on the surface of photosensitive material of work function $$\displaystyle h\nu_{0},$$ then

A
The kinetic energy of all emitted electrons is $$\displaystyle hv_{0}$$

B
The kinetic energy of all emitted electrons is $$\displaystyle h( v-v_{0})$$

C
The kinetic energy of all fastest electrons is $$\displaystyle h( v-v_{0})$$

D
The kinetic energy of all emitted electrons is $$hv$$

Solution

The maximum kinetic energy ($$K_{max}$$) of the photoelectrons to the frequency of the absorbed photons ($$\nu$$) and the threshold frequency ($$\nu_o$$) of the photoemissive surface.
$$K_{max} = h (\nu - \nu_o)$$
So, the answer is option (C).
Question 5
1 / -0

The work function of cesium metal is 2eV. It means that

A
The energy necessary to emit electrons from metal surface is 2eV

B
The energy of electrons emitted from metallic surface is 2eV

C
The value of photoelectric current is 2eV

D
The value of threshold frequency is 2eV

Solution

By definition the work function is the minimum energy (in electron volts) needed to remove an electron from a solid to a point immediately outside the solid surface (or energy needed to move an electron from the Fermi level into vacuum).
Hence, if the work function of cesium metal is 2ev, it means that the energy necessary to emit electrons from metal surface is 2eV.
So, the answer is option (A).
Question 6
1 / -0

The effective mass of photon of wavelength 40$$ \overset { \circ }{ A } $$ will be:

A
$$ 55.2\times { 10 }^{ -35 }kg$$

B
$$ 55.2\times { 10 }^{ -33 }gm$$

C
$$ 55.2\times { 10 }^{ -17 }kg$$

D
$$ 55.2\times { 10 }^{ -38 }kg$$

Solution

From de Broglie hypothesis, wavelength of photon , $$\lambda=\dfrac{h}{p}=\dfrac{h}{mc}$$ where $$h=$$ Planck's constant , $$m=$$ effective mass and $$c=$$ velocity of light or photon.
So, $$m=\dfrac{h}{\lambda c}=\dfrac{6.62\times 10^{-34}}{(40\times 10^{-10})\times (3\times 10^8)}=55.2\times 10^{-35} kg$$
Question 7
1 / -0

Which conservation law is obeyed in Einstein's photoelectric equation?

A
Charge

B
Energy

C
Momentum

D
Mass

Solution

Einstein's photoelectric effect obeys law of conservation of energy.
So, the answer is option (B).
Question 8
1 / -0

Find the minimum wavelength of X-ray produced if 10 kV potential difference is applied across the anode and cathode of the tube.

A
12.4 A$$^{\circ}$$

B
12.4 nm

C
1.24 nm

D
1.24 A$$^{\circ}$$

Solution

If electrons are accelerated to a velocity $$v$$ by a potential difference V and then allowed to collide with a metal target, the maximum frequency of the X-rays emitted is given by the equation:
$$ mv^2= eV = h\nu $$

or, $$\displaystyle \lambda_{min} = \dfrac{1240}{10^3}*10^{-9} = 124 nm = 12.4 A^{\circ} $$

So, the answer is option (D).
Question 9
1 / -0

The photoelectric equation is

A
$$\displaystyle hv= hv_{0}-E_{k}$$

B
$$\displaystyle kv= hv_{0}+\frac{E_{k}}{V}$$

C
$$\displaystyle hv= hv_{0}+E_{k}$$

D
$$\displaystyle hv= hv_{0}$$

Solution

The photoelectric effect related with the
incident light frequency $$(\nu)$$ by the following equation:
$$E_k=eV_s=h\nu-\phi$$,
where $$\phi=h\nu_0$$ is the work function of the material and $$E_k$$ is the kinetic energy of the photo electron.
So, $$E_k=h\nu-h\nu_0$$
or, $$h\nu=h\nu_0+E_k$$
So, the answer is option (C).
Question 10
1 / -0

Einstein was awarded Noble prize for -

A
The general theory of relativity

B
The special theory of relativity

C
The explanation of photoelectric effect

D
The explanation of quantum theory

Solution

Einstein proposed the special theory of relativity and the general theory of relativity. Also, he gave a explanation to quantum theory. According to Nobel committee other theories were in controversy.
The classical theory could not explain photoelectric effect satisfactorily. The photoelectric equation given by Einstein explains photoelectric effect satisfactorily.
Hence, Einstein was awarded Noble prize for the explanation of photoelectric effect.
So, the answer is option (C).

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Re-Attempt Weekly Quiz Competition

Dual Nature of Radiation and Matter Test - 29

Result

Dual Nature of Radiation and Matter Test - 29

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SHARING IS CARING

Question 1

28.6 $$pm$$

2.86 $$pm$$

2.86 $$fm$$

28.6 $$fm$$

Solution

Question 2

207 $$nm$$

702 $$nm$$

207 $$A^{\circ}$$

2070 $$nm$$

Solution

Question 3

3 $$A^{\circ}$$

3 $$pm$$

3 $$fm$$

30 $$fm$$

Solution

Question 4

The kinetic energy of all emitted electrons is $$\displaystyle hv_{0}$$

The kinetic energy of all emitted electrons is $$\displaystyle h( v-v_{0})$$

The kinetic energy of all fastest electrons is $$\displaystyle h( v-v_{0})$$

The kinetic energy of all emitted electrons is $$hv$$

Solution

Question 5

The energy necessary to emit electrons from metal surface is 2eV

The energy of electrons emitted from metallic surface is 2eV

The value of photoelectric current is 2eV

The value of threshold frequency is 2eV

Solution

Question 6

$$ 55.2\times { 10 }^{ -35 }kg$$

$$ 55.2\times { 10 }^{ -33 }gm$$

$$ 55.2\times { 10 }^{ -17 }kg$$

$$ 55.2\times { 10 }^{ -38 }kg$$

Solution

Question 7

Charge

Energy

Momentum

Mass

Solution

Question 8

12.4 A$$^{\circ}$$

12.4 nm

1.24 nm

1.24 A$$^{\circ}$$

Solution

Question 9

$$\displaystyle hv= hv_{0}-E_{k}$$

$$\displaystyle kv= hv_{0}+\frac{E_{k}}{V}$$

$$\displaystyle hv= hv_{0}+E_{k}$$

$$\displaystyle hv= hv_{0}$$

Solution

Question 10

The general theory of relativity

The special theory of relativity

The explanation of photoelectric effect

The explanation of quantum theory

Solution

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