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

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

A

`1.2 xx 10^(-18)J`

B

`1.2 xx 10^(-20)J`

C

`6 xx 10^(-19)J`

D

`6 xx 10^(-12)J`

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The correct Answer is:

C

Work function is `hv_(0) = (hc)/(lambda_(0))`
`= (6.62 xx 10^(-34) xx 3 xx 10^(8))/(330 xx 10^(-9)) = 6 xx 10^(-19)J`.

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