Question

Electromagnetic radiation having $\lambda =310\stackrel{\circ }{\text{A}}$ is subjected to a metal sheet having work function $=12.8\text{eV}$. What will be the velocity of photoelectrons having maximum kinetic energy.

• A. $0,$ no emission will occur
• B. $4.352\times {10}^8\text{m/s}$
• C. $3.09\times {10}^6\text{m/s}$
• D. $8.72\times {10}^6\text{m/s}$

Answer 1

The correct option is C $3.09\times {10}^6\text{m/s}$

Given: $\lambda =310\stackrel{\circ }{\text{A}}$
$W_o=12.8\text{eV}=2.048\times {10}^{-18}J$
Using the Einstein's Photoelectric Equation,
$KE=E-W_o$
$\frac{1}{2}m{\text{v}}^2=\frac{hc}{\lambda }-W_o$

$\Rightarrow \frac{(6.626\times {10}^{-34}\text{J s})(3\times {10}^8\text{m/s})}{310\times {10}^{-10}\text{m}}=\left(12.8\text{eV}\right)(1.6\times {10}^{-19}\text{J/eV})+\frac{1}{2}m{\text{v}}^2$

$\Rightarrow 0.0641\times {10}^{-16}=(2.048\times {10}^{-18}\text{J})+\frac{1}{2}m{\text{v}}^2$

$\Rightarrow (64.1\times {10}^{-19}\text{J}-20.48\times {10}^{-19}\text{J})=\frac{1}{2}m{\text{v}}^2$

$\Rightarrow \frac{2\times 43.62\times {10}^{-19}\text{J}}{9\times {10}^{-31}\text{kg}}={\text{v}}^2[\because m_e=9\times {10}^{-31}\text{kg}]$

$\Rightarrow \text{v}=3.09\times {10}^6\text{m/s}$