Question

When radiation is incident on a photoelectron emitter, the stopping potential is found to be 9 volts. If e/m for the electron is $1.8 \times 10^{11}Ckg^{- 1}$ the maximum velocity of the ejected electrons is

• A. $6 \times 10^{5}ms^{- 1}$
• B. $8 \times 10^{5}ms^{- 1}$
• C. $1.8 \times 10^{6}ms^{- 1}$
• D. $1.8 \times 10^{5}ms^{- 1}$

Answer 1

Answer: (C)

$1.8 \times 10^{6}ms^{- 1}$

Answer 2

$\frac{1}{2}mv_{\max}^{2_{0}}$⇒

${v\sqrt{2\left( \frac{e}{m} \right).V_{0}}}_{\max}$

$= \sqrt{2 \times 1.8 \times 10^{11} \times 9} = 1.8 \times 10^{6}m/s$