Question

The photoelectric threshold of Tungsten is 2300$\mathring{A}$. The energy of the electrons ejected from the surface by ultraviolet light of wavelength 18000 $\mathring{A}$ is

• A. 0.15 e V
• B. 1.5 e V
• C. 15 e V
• D. 150 e V

Answer 1

The correct option is A 0.15 e V

$E_k=d\frac{h}{c}(\frac{1}{\lambda }-\frac{1}{{\lambda }_o})\left(ineV\right)$
$=\frac{6.6\times {10}^{34}\times 3\times {10}^8}{1.6\times {10}^{-19}}$$(\frac{{10}^{10}}{1800}-\frac{{10}^{10}}{2300})=0.15eV$