Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency `(V_(0))` and work function `(W_(0))` of metal.

Explanation

Given `= 6800 Å = 6800 xx 10^(-8) cm = 6.8 xx 10^(-5) cm`
`:. v_(0) =(c)/(lambda) = (3 xx 10^(10) cm.sec^(-1))/(6.8 xx 10^(-5) cm) = 0.44 xx 10^(15) s^(-1)`
`= 4.4 xx 10^(14) s^(-1)` (or Hz)
`W_(0) = h lambda_(0) = 6.625 xx 10^(-27)` erg.sec `xx 4.4 xx 10^(14) S^(-1)`.
`= 29.15 xx 10^(-13)` erg. for 1 photo electrons.