When a certain metal was irradiated with light of frequency `3.2 xx 10^(16)s^(-1)` the photoelectrons emitted had twice the KE as did photoelectrons emitted when the same metal was irradiated with light of frequency`2.0 xx 10^(16)s^(-1)` . Calculate the thereshold frequency of the metal.

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`