1.

For given metal work function is W_(0) and threshold wavelength is lambda_(0).If metal has work function (W_(0))/(2) then what will be threshold wavelength ?

Answer»

`(lambda_(0))/(4)`
`(lambda_(0))/(2)`
`2lambda_(0)`
`4lambda_(0)`

Solution :`W_(0)=(HC)/(lambda_(0))` and `W_(0)=(hc)/(lambda_(0))`
`THEREFORE (W_(0))/(W_(0))=(lambda_(0))/(lambda_(0))`
`(W_(0))/(W_(0))=(lambda_(0))/(lambda_(0))`
`(W_(0))/((W_(0))//(2))=(lambda_(0))/(lambda_(0)) [because W_(0)=(W_(0))/(2)]`
`therefore 2lambda_(0)=lambda_(0)`


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