Question

When the electromagnetic radiations of frequencies `4 xx 10^(15) Hz` and `6 xx 10^(15) Hz` fall on the same metal, in different experiments, the ratio of maximum kinetic energy of electrons liberated is `1 : 3`. The threshold frequency for the metal is
A. `2 xx 10^(15)` Hz
B. `1 xx 10^(15)` Hz
C. `3 xx 10^(15)` Hz
D. `1.67 xx 10^(15)` Hz

Answer 1

Correct Answer - C
`hv=hv_(0)+KE_("max")`
For first experiment,
`h xx4 xx 10^(15)=hv_(0)+x" "….(i)`
For second experiment,
`h xx 6 xx 10^(15)=hv_(0)+3x" ....(ii)"`
From Eqs. (i) and (ii), we get `v_(0)=3xx10^(15)Hz`