When radiation of wavelength λ is incident on a metallic surface, the stopping potential of ejected photoelectrons is 4.8 V. If the same surface is illuminated by radiation of double the previous wavelength, then the stopping potential becomes 1.6 V. The threshold wavelength of the metal is:
When radiation of wavelength λ is incident on a metallic surface, the stopping potential of ejected photoelectrons is 4.8 V. If the same surface is illuminated by radiation of double the previous wavelength, then the stopping potential becomes 1.6 V. The threshold wavelength of the metal is:
Option 1 -
6λ
Option 2 -
8λ
Option 3 -
2λ
Option 4 -
4λ
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1 Answer
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Correct Option - 4
Detailed Solution:By Einstein's equation of photoelectric effect,
hc/λ = hc/λ? + eVs
Now hc/λ = hc/λ? + e×4.8 . (i)
hc/ (2λ) = hc/λ? + e×1.6 . (ii)
Multiply (ii) by 2:
hc/λ = 2hc/λ? + e×3.2 . (iii)
Equating (i) and (iii):
hc/λ? + 4.8e = 2hc/λ? + 3.2e
1.6e = hc/λ?
Substitute this into (i)
hc/λ = 1.6e + 4.8e = 6.4e
Substitute this into (ii)
(6.4e)/2 = hc/λ? + 1.6e
3.2e = hc/λ? + 1.6e
1.6e = hc/λ?
From (i): hc/λ = hc/λ? + 3 (hc/λ? ) = 4hc/λ?
λ? = 4λ
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