Question

The work function of a metal is $2eV$. If a radiation of wavelength $3000$ $\mathring{A}$ is incident on it, the maximum kinetic energy of the emitted photoelectrons is :
(Planck's constant $h=6.6\times {10}^{-34}Js$; Velocity of light $c=3\times {10}^{8}m/s;1eV=1.6\times {10}^{-19}J$)

• A. $4.4\times {10}^{-19}J$
• B. $5.6\times {10}^{-19}J$
• C. $3.4\times {10}^{-19}J$
• D. $2.5\times {10}^{-19}J$

Answer 1

The correct option is C $3.4\times {10}^{-19}J$

Energy of incident radiation is $E=\frac{hc}{\lambda }=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{3\times {10}^{-7}}=6.6\times {10}^{-19}J$.

Also, work function is $\varphi =2\times 1.6\times {10}^{-19}J=3.2\times {10}^{-19}J$.

Hence, maximum kinetic energy of emitted photo electrons will be $6.6\times {10}^{-19}J-3.2\times {10}^{-19}J=3.4\times {10}^{-19}J$.