Question

In the photoelectric effect, the maximum speed of electrons is found to be $6\times 10^{5} \, m \, s^{- 1}$ . The wavelength used is $4000 \, Å$ . The work function of the metal is

• A. $\text{2.2}eV$
• B. $\text{2.076}eV$
• C. $\text{2.3}eV$
• D. $\text{2.4}eV$

Answer 1

Answer: (B)

$\frac{hc}{\lambda }=\left(KE\right)_{Max}+\phi$
$\frac{12400}{4000}=\frac{\frac{1}{2} \times 9 .1 \times 10^{- 31} \times 36 \times 10^{10}}{1 .6 \times 10^{- 19}}+\phi$
$\text{3.1}=\text{102.375}\times 10^{- 2}+\phi$
$\phi=\text{2.076}eV$