Light of wavelength λ strikes a photoelectric surface and electrons are ejected with kinetic energy K. If K is to be increased to exactly twice its original value, the wavelength must be changed is λ′ such that:
λ′<λ/2
λ′>λ/2
λ>λ′>λ/2
λ′=λ/2
A
λ′<λ/2
B
λ′>λ/2
C
λ>λ′>λ/2
D
λ′=λ/2
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Solution
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K=hcλ−ϕ
where ϕ is the work potential.
ϕ=hcλ−K
Now to increase K to 2K
2K=2hcλ′−2ϕ
Replacing ϕ
2K=2hcλ′−2hcλ+2K
λ′=λλ+1
Thus, λ′<λ. Hence option C is correct.
Also since we are talking about particle nature of light, λ is small,
thus λ′>λ2.
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