Question

Maximum velocity of photoelectron emitted is $4.8\, ms ^{-1}$. The $\frac{e}{M}$ ratio of electron is $1.76 \times 10^{11} Ckg ^{-1}$, then stopping potential is given by

• A. $5 \times 10^{-10} JC ^{-1}$
• B. $3 \times 10^{-7} JC ^{-1}$
• C. $7 \times 10^{11} JC ^{-1}$
• D. $2.5 \times 10^{-2} JC ^{-1}$

Answer 1

Answer: (C)

As, $eV _{s}=\frac{1}{2} m v_{m}^{2}$
or $V_{s}=\frac{m v_{m}^{2}}{2 e}=\frac{v_{m}^{2}}{2(e / m)}$
$=\frac{(4.8)^{2}}{2 \times 1.76 \times 10^{-11}}=7 \times 10^{11} JC ^{-1}$