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Q.

A metal is irradiated by two incident radiations in two different experiments. When the ratio of incident frequencies is 1:2 then the ratio of kinetic energy of photo electrons is 1:3. Then the correct relation is(ν1, ν2 are incident frequencies in experiment one and two respectively and ν0 is the threshold frequency)

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a

3ν0=ν1

b

2ν0=ν1

c

ν0=ν1

d

1.5ν0=ν1

answer is B.

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Detailed Solution

For the same metal, work function(W) remains constant;hν1=hν0+K.E1....1 hv2=hν0+K.E2.....(2)      K.E1K.E2=ν1-v0ν2-v0;           13=ν1-ν02ν1-ν0;  2ν1-ν0=3(ν1-ν0)         2ν0=ν1
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