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Test: Photoelectric Effect


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Test: Photoelectric Effect - Question 1

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

Detailed Solution for Test: Photoelectric Effect - Question 1

Kmax=eV0=hf-hf0
Kmax=eV0= h(f-f0)

Test: Photoelectric Effect - Question 2

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​

Detailed Solution for Test: Photoelectric Effect - Question 2

The maximum kinetic energy of the emitted electrons is given by
Kmax​=hυ−ϕ0​=h(4υ)−h(υ)=3hυ

Test: Photoelectric Effect - Question 3

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

Detailed Solution for Test: Photoelectric Effect - Question 3

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

Test: Photoelectric Effect - Question 4

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​

Detailed Solution for Test: Photoelectric Effect - Question 4

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.

Test: Photoelectric Effect - Question 5

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

Detailed Solution for Test: Photoelectric Effect - Question 5

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
 

Test: Photoelectric Effect - Question 6

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

Detailed Solution for Test: Photoelectric Effect - Question 6

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.

Test: Photoelectric Effect - Question 7

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

Detailed Solution for Test: Photoelectric Effect - Question 7

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 .
 

Test: Photoelectric Effect - Question 8

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?​

Detailed Solution for Test: Photoelectric Effect - Question 8

Stopping potential (Vo​) is given by
Vo​=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, Vo​=(10eV)/e​=10V

Test: Photoelectric Effect - Question 9

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​

Detailed Solution for Test: Photoelectric Effect - Question 9

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

Test: Photoelectric Effect - Question 10

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

Detailed Solution for Test: Photoelectric Effect - Question 10

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

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