Question

When radiation is incident on a photoelectron emitter, the stopping potential is found to be $9V$. If elm for the electron is $1.8 \times 10^{11} \, C \, kg^{-1}$, the maximum velocity of the ejected electron 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

$\frac{1}{2} mv _{\max }^{2}= eV _{0}$
$\Rightarrow v _{\max }=\sqrt{2 \frac{ e }{ m } V _{0}}$
$=\sqrt{2 \times 1.8 \times 10^{11 \times 9}}$
$=18 \times 10^{5} \,m / s$
$=1.8 \times 10^{6} \,m / s$