Question

If the threshold wavelength $\left({\lambda }_0\right)$ for ejection of electron from metal is 330 nm, then work function for the photoelectric emission is:
(Planck's constant (h) = $6.6\times {10}^{-34}\text{J s}$)

• A. $1.2\times {10}^{-18}\text{J}$
• B. $1.2\times {10}^{-20}\text{J}$
• C. $6\times {10}^{-19}\text{J}$
• D. $6\times {10}^{-12}\text{J}$

Answer 1

The correct option is C $6\times {10}^{-19}\text{J}$

From the Einstein photoelectric effect, work function is the minimum energy required to remove an electron from the metal surface.
Work function $=$ Threshold energy
$⟹h{\nu }_0=\frac{hc}{{\lambda }_0}=\frac{6.6\times {10}^{-34}\text{J s}\times 3\times {10}^8\text{m/s}}{330\times {10}^{-9}\text{m}}=6\times {10}^{-19}\text{J}$