Question

Determine the maximum kinetic energy of emitted photo electrons.

• A. $1.80eV$
• B. $1.08eV$
• C. $8.0eV$
• D. $0eV$

Answer 1

Answer: (A)

$1.80eV$

Wavelength of light $\lambda =365\times {10}^{-9}m$
Energy of one photon $E=\frac{hc}{\lambda }J=\frac{hc}{e\lambda }eV$
$\therefore$ $E=\frac{(6.62\times {10}^{-34})(3\times {10}^8)}{(1.6\times {10}^{-19})(365\times {10}^{-9})}=3.4eV$
Work function of surface $\varphi =1.6eV$
Thus maximum kinetic energy of photo electron $K.E_{max}=E-\varphi$
$⟹K.E_{max}=3.4-1.6=1.80eV$