1.

When a metallic surface is illuminated with radiation of wavelength lambda, the stoppingpotentials 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

Answer»

`4lambda`
`5 lambda`
`(5)/(2) lambda`
`3 lambda`

Solution :`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 solvingwe GET`lambda_(0) = 3 lambda`


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