The threshold wavelength for certain metal is `lambda _(0)`. When a light of wavelength ` (lambda_(0))//(2) ` is incident on it , the mximum velocity of photelectrons is `10^(6) m//s`. If the wavelength of the incident radiation is reduced to `(lambda _(0) // (5)`, then the maximum velocity of the photoelectrons in `m//s` will be ,

Answer

2

Explanation

`(lambda_1)/(lambda_2) = sqrt((T_2)/(T_1))= sqrt((1200)/(300)) = 2 `