A certain metal when irradiated by light `(v=3.2xx10^(16)Hz)` emits photoelectrons with twice of K.E. as did photoelectrons when the same metal is irradiated by light `(v=2.0xx10^(16)Hz)`. The `v_(0)` of the metal is

Answer

`319.2kJ//mol`

Explanation

`hv_(1)=hv_(0)+2E_(1)" "hv_(2)=hv_(0)+E_(1)`
`hv_(1)-w_(0)+2E_(1)" "hv_(2)-w_(0)+E_(1)`
`2=(hv_(1)-w_(0))/(hv_(2)-w_(0))" "2 hv_(2)-2w_(0)=hv_(1)-w_(0)`
`h[2v_(2)-v_(1)]=w_(0)`
`w_(0)=6.62xx10^(-34)(2xx10^(15)-3.2xx10^(15))`
`w_(0)=6.62xx10^(-34)xx0.8xx10^(15)`
`w_(0)=5.29xx10^(-19)" "w_(0)=318.9kJ//mol`