Question

Find the work function of metal.

• A. $2eV$
• B. $2.8eV$
• C. $0.78eV$
• D. $1.03eV$

Answer 1

The correct option is D $1.03eV$

Maximum kinetic energy of photoelectron emitted $K=0.086eV$ (solved already)
Wavelength of incident photon $\frac{1}{\lambda }=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
$\therefore$ $\frac{1}{\lambda }=(1.097\times {10}^7)(\frac{1}{2^2}-\frac{1}{3^2})$
$⟹\lambda =6.567\times {10}^{-7}m=6567A^o$
Energy of incident photon $E=\frac{hc}{\lambda }=\frac{(6.62\times {10}^{-34})(3\times {10}^8)}{6567\times {10}^{-10}}J=3.024\times {10}^{-19}J$
$⟹E=\frac{3.024\times {10}^{-19}}{1.6\times {10}^{-19}}=1.89eV$
Thus work function of the metal $\varphi =E-K=1.89-0.86=1.03eV$