1.

When a metallic surface is illuminated with radiation of wavelength lambda, the stopping potential 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»

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

Solution :`eV = (hc)/(lambda) - (hc)/(lambda_(0))""…(1)""[K_(max) = h upsilon - phi_(0)]`
`E((V)/(4)) = (hc)/(2 lambda) - (hc)/(lambda_(0))""…(2)`
`(1) DIV (2)`
`(4 (eV))/(eV) = ((1)/(lambda) - (1)/(lambda_(0)))/((1)/(2 lambda) - (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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