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When the electromagnetic radiations of f...

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

Text Solution

Verified by Experts

The correct Answer is:
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`
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