If threshold wavelength (lambda(0)) for ...

If threshold wavelength `(lambda_(0))` for ejection of electron from metal is 330 nm, then work function for the photoelectron emission is

`1.2xx10^(-18) J`

`6.0xx10^(-19) J`

`1.2xx10^(-20) J`

`6.0xx10^(-12) J`

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The correct Answer is:

B, D

Photoelectric work function `=hv_(0)=(hc)/(lambda)`
`=(6.62xx10^(-34)xx3xx10^(8))/(330xx10^(-9))=6xx10^(-19) J`

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