Question

Light of wavelength $4000 \, Å$ is allowed to fall on a metal surface having work function $2 \, eV$ . The maximum velocity of the emitted electrons is $\left(\right.h=6.6\times \left(10\right)^{- 34} \, Js$ )

• A. $1.35\times 10^{5} \, m \, s^{- 1}$
• B. $2.7\times 10^{5} \, m \, s^{- 1}$
• C. $6.2\times 10^{5} \, m \, s^{- 1}$
• D. $8.1\times 10^{5} \, m \, s^{- 1}$

Answer 1

Answer: (C)

$\frac{1}{2}mv^{2}=\frac{h c}{\lambda }-\phi \, \left(\right.in \, eV\left.\right)$
$=\frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{4000 \times 10^{- 10} \times 1.6 \times 10^{- 19}}-2$
$=3.1-2=1.1 \, eV=1.1\times 1.6\times 10^{- 19} \, J$
$=1.76\times 10^{- 19} \, J$
$v=\frac{1.76 \times 10^{- 19} \times 2}{9 \times 10^{-3 1}}$
$=6.2\times 10^{5} \, m \, s^{- 1}$