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Q.

Light of wavelength 4000 A∘  is incident on a potassium surface whose work function is 2.1 ev . What will be the stopping potential

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a

1 v

b

2.1 v

c

4000 v

d

3.1 v

answer is A.

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Detailed Solution

Wavelength λ=4000 A∘ Work function w=2.1 ev E=w+ev0 12,4004000=2.1+ev0 (∵E=12400λ A∘) 3.1=2.1+ev0   v0=1 v
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