Question

Maximum velocity of photoelectrons emitted by a photo emitter is $1.8 \times 10^{6} m / s$. Taking $e / m=1.8 \times 10^{11}$ for electrons, the stopping potential of the emitter is

• A. 9 V
• B. 11.8 V
• C. 1.8 V
• D. 10 V

Answer 1

Answer: (A)

Stop pot $( Vs )$ in Volt $= KE ( eV )$
$=1 / 2\, m V ^{2} J$
$=1 / 2\left( m V ^{2}\right) / e$
$eV =1 / 2 V ^{2} /( e / m )=1 / 2 \times\left[\left(1.8 \times 10^{6}\right)^{2}\right] /1.8 \times 10^{11}=9\, eV$
$Vs =9\, V$