electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 A . Calculate thrshold frequency `(v_(0))` and work funcation `(W_(0))` of the metal.

Explanation

`hv=W+1/2 mv^2`
Since electrons are emitted with zero velocity v=0
Hence `hv=W_(0)`
`W_(0)=hv=(hc)/(lambda)=(6.626 xx 10^(-34) xx 3 xx 10^(8))/(6800 xx 10^(-10))`
`=.2.923 xx 10^(-19)J`
If `v_(0)` is the threshold frequency `W=hv_(0)`
`or v_(0)=W/h=(2.923 xx 10^(-19))/(6.626 xx 10^(-34)) =4.412 xx 10^(14) s^(-1)`