The threshould frequency for a metallic surface corresponds to an energy of `6.2eV` and the stopping potential for a radiation insident on this surface is `5 V` . The incident radiation lies in
A
ultra - violet region
B
infra- red regaion
C
visible region
D
X- ray ragion
Text Solution
Verified by Experts
The correct Answer is:
A
`phi = 6.2 eV= 6.2 xx 10^(-19)J` `rArr lambda = (hc)/(phi + eV_(0)) = (6.6 xx 10^(-34) xx 3 xx 10^(8))/(1.6 xx 10^(-19) (6.2 + 5))= 10^(-7) m ` This rang lies in ultra violet range
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