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When a metallic surface is illuminated w...

When a metallic surface is illuminated with radiation of wavelength `lambda`, the stopping potentials is V. If the same surface is illuminated with radiation of wavelength `2 lambda`, the stopping potential is `( V )/( 4)`. The threshold wavelength for the metallic surface is

A

`4 lambda`

B

`5 lambda`

C

`(5)/(2) lambda`

D

`3 lambda`

Text Solution

Verified by Experts

The correct Answer is:
D
`eV = (hc)/(lambda) - (hc)/(lambda_(0)) " "…….(1)` `" "[K_("max") = hv - phi_(0)]`
`e((V)/(4)) = (hc)/(2lambda) - (hc)/(lambda_(0))" "……(2)`
`(1)+ (2)`
`(4(eV))/(eV)` = `((1)/(lambda) - (1)/(lambda_(0)))/((1)/(2lambda) - (1)/(lambda_(0))) = (((lambda_(0) - lambda)/(lambda lambda_(0))))/(((lambda_(0) - 2 lambda)/(2 lambda lambda_(0))))`
On solving we get `lambda_(0) = 3 lambda`
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