1.

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»


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


Discussion

No Comment Found