Dual Nature of Radiation and Matter Test 2

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Dual Nature of Radiation and Matter Test 2

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Weekly Quiz Competition

Question 1
1 / -0

Photons of frequency f is incident on a metal surface of threshold frequency f₀. The maximum K.E. of emitted photoelectrons is

A
h(f + f₀)

B
hf₀

C
h(f – f₀)

D
hf

Solution

K_max=eV₀=hf-hf₀
K_max=eV₀= h(f-f₀)
Question 2
1 / -0

The photoelectric threshold frequency of a metal is f₀. When light of frequency 4f₀is incident on the metal, the maximum K.E. of the emitted electron is

A
2 hf₀

B
4 hf₀

C
3 hf₀

D
hf₀/ 2

Solution

The maximum kinetic energy of the emitted electrons is given by
K_max=hυ−ϕ₀=h(4υ)−h(υ)=3hυ
Question 3
1 / -0

Sodium surface is illuminated by ultraviolet and visible radiation successively and the stopping potential determined. The stopping potential is

A
Greater with visible light

B
Equal in both cases

C
Greater with ultraviolet light

D
Infinite in both cases

Solution

λ for U.V is less than λ for visible light
ν for U.V is greater than ν for visible light
∴ potential is greater for U.V light.
as K.Eα1/λ
Question 4
1 / -0

Light of frequency 1.5 times the threshold frequency is incident on a photosensitive material .If the frequency is halved and the intensity is doubled, the photoelectric current becomes

A
Doubled

B
Quadrupled

C
Halved

D
zero

Solution

When the frequency of incident light is halved of its original value i.e. 1.5v0 then it becomes less than the threshold value. In that case no photoelectric effect takes place. No photoelectrons would be emitted. The photoelectric current becomes zero.
Question 5
1 / -0

Light of wavelength 4000 Â is incident on a metal of work function 3.2 x 10^-19 J. What is the maximum kinetic energy of the emitted electron?

A
2.2 x 10^-19 J

B
1.75 x 10^-19 J

C
0.75 x 10^-19 J

D
1.1 x 10^-19 J

Solution

Given,
λ= 4000 Å ;W•= 3.2× 10-19 j
as we know that
K.E = (hc/ λ) - W•
= (6.6× 10-34 × 3 × 10-19/ 4000×10-10 )- 3.20× 10^-19
= 1.75 × 10^-19
Question 6
1 / -0

Light from a bulb is falling on a wooden table but no photo electrons are emitted as

A
Work function of wood is less

B
Work function of wood is more

C
It depends on the frequency

D
It is independent of work function

Solution

It is because the work function of wood is more than the energy of the incident photons of light. Hence, photoelectrons are not emitted from the wooden table.
Question 7
1 / -0

Photons of energy 6eV are incident on a potassium surface of a work function 2.1 eV. What is the stopping potential?

A
-3.9V

B
-8.1V

C
-5V

D
-1.9V

Solution

From photo-electric equation, eV₀= E−φ
eV₀=(6−2.1)eV
V₀= 3.9 V
stopping potential is a negative potential to stop e- at saturated current .
Question 8
1 / -0

The energy of the incident photon is 20 eV and the work function of the photosensitive metal is 10 eV. What is the stopping potential?

A
30 V

B
5 V

C
10 V

D
15 V

Solution

Stopping potential (V_o) is given by
V_o=W/q where W is the work function and q are the charge of an electron.
Given W=20eV−10eV=10eV. Also, q=e
Hence, V_o=(10eV)/e=10V
Question 9
1 / -0

Light of two different frequencies whose photons have energies of 1eV and 2.5 eV respectively successively illuminate a metal of work function 0.5eV. The ratio of maximum speed of emitted electrons is

A
1:5

B
1:4

C
1:3

D
1:2

Solution

For maximum speed of the photo electrons (½)mv2=Ephoton−ϕ
where ϕ=0.5 eV is the work function of the metal.
Energy of photon E1=1eV
∴ (½)mv12=1−0.5
We get (½) mv12=0.5 eV ....(1)
Energy of photon E₂=2.5eV
∴ (½) mv₂²=2.5−0.5
We get (½)mv₂²=2 eV ....(2)
Ratio of maximum speeds v12/ v22=0.5/2=1/4
We get v₁/v₂ =1/2
Question 10
1 / -0

A photo-sensitive material would emit electrons, if excited by photons beyond a threshold. To overcome the threshold, one would increase the:

A
voltage applied to the light source

B
intensity of light

C
wavelength of light

D
frequency of light

Solution

By theory
for photoelectric effect,
hf=hf₀+E_k, Here f>f₀

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Dual Nature of Radiation and Matter Test 2

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Dual Nature of Radiation and Matter Test 2

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Question 1

h(f + f0)

hf0

h(f – f0)

hf

Solution

Question 2

2 hf0

4 hf0

3 hf0

hf0/ 2

Solution

Question 3

Greater with visible light

Equal in both cases

Greater with ultraviolet light

Infinite in both cases

Solution

Question 4

Doubled

Quadrupled

Halved

zero

Solution

Question 5

2.2 x 10-19 J

1.75 x 10-19 J

0.75 x 10-19 J

1.1 x 10-19 J

Solution

Question 6

Work function of wood is less

Work function of wood is more

It depends on the frequency

It is independent of work function

Solution

Question 7

-3.9V

-8.1V

-5V

-1.9V

Solution

Question 8

30 V

5 V

10 V

15 V

Solution

Question 9

1:5

1:4

1:3

1:2

Solution

Question 10

voltage applied to the light source

intensity of light

wavelength of light

frequency of light

Solution

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h(f + f₀)

hf₀

h(f – f₀)

2 hf₀

4 hf₀

3 hf₀

hf₀/ 2

2.2 x 10^-19 J

1.75 x 10^-19 J

0.75 x 10^-19 J

1.1 x 10^-19 J