When light of wavelength 300 nm (manometer) falls on a photo electric emitter, photo electrons are liberated. For another emitter, however, light of 600 nm wavelength is sufficient for creating photo-emission. What is the ratio of the work-functions of the two emitter ?
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a
1:2
b
2 :1
c
4 : 1
d
1 :4
answer is B.
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Detailed Solution
if λ0 be the threshold wavelength, thenW=hc/λ0 W1W2=λ2λ1=21