1.

The threshold wavelength for the ejection of electron from metal is 330 nm, then work function for the photoelectric emission is

Answer»

`6XX10^(-12)` J
`6xx10^(-19)` J
`1.2xx10^(-20)` J
`1.2xx10^(-18)` J

Solution :`W=hv_(0)`
`W=(hc)/lambda_(0)=(6.62xx10^(-34)xx3xx10^(8))/(330xx10^(-9))=6xx10^(-19)` J


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