If the threshold wavelength `(lambda_(0))` for spection of electron from metal is `350nm `then work function for the photoelectric emission is

Options

A. `1.2 xx 10^(-18) J`
B. `1.2 xx 10^(-20) J`
C. `6 xx 10^(-29) J`
D. `6 xx 10^(-12) J`

Answer

B

Explanation

Work function = Threshold energy
`= hv_(0) = (hc)/(lambda_(0))`
` = (6.6 xx 10^(-34) J s xx 3 xx 10^(8) m)/(330 xx 10^(-9) m) = 6.6 xx 10^(-29) J`