Two identical photo cathodes receive light of frequencies f1 and f2 If the velocities of the photo electrons (of mass m) coming out are respectively v1 and v2,then
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a
v12−v22=2hmv1−v2
b
v1+v2=2hmv1+v21/2
c
v12+v22=2hmv1+v2
d
v1−v2=2hmv1−v21/2
answer is A.
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Detailed Solution
hv=hv0+12mv2 v12=2hv1m−2hv0m and v22=2hv2m−2hv0m v12−v22=2hmv1−v2