Light rays of wavelength `lambda and (lambda)/(2)` are incident on a photosensitive metal surface. If the maximum kinetic energy of the emitted photoelectrons from the metal surface in 2nd case be 3 times the maximum kinetic energy of emitted photoelectrons in the 1st case, then determine the work function of the metal.

Explanation

`E_("max") = hf - W_(0) = (hc)/(lambda) - W_(0)`
1st case, `E_("max") = (hc)/(lambda) -W_(0)`
2nd case, `3E_("max") = (hc)/(lambda//2) - W_(0) = (2hc)/(lambda) - W_(0)`
or, ` " " E_("max") =(2)/(3) (hc)/(lambda)-(1)/(3) W_(0)`
`therefore " " (hc)/(lambda) - W_(0) = (2)/(3) (hc)/(lambda)-(1)/(3) W_(0)`
or, `" " (hc)/(lambda)(1-(2)/(3)) = W_(0) (1 - (1)/(3)) or, W_(0) = ((1)/(3)(hc)/(lambda))/(2//3) = (1)/(2)(hc)/(lambda)`