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 "= 6800Å = 6800 xx 10^(-8) cm`
`= 6.8 xx10^(-5) cm`
`therefore v_(0)=(c)/(lambda)=(3xx10^(10)" cm. sec"^(-1))/(6.8xx10^(-5)cm)=0.44xx10^(15)s^(-1)`
`=4.4xx10^(14)s^(-1)" (or Hz)"`
`W_(0)=h lambda_(0)=6.625xx10^(-27)" erg. Sec" xx4.4xx10^(14)S^(-1)`.
`29.15xx10^(-13)" erg. For 1 photo electron."`