Question

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

• A. $9V$
• B. $11.8V$
• C. $1.8V$
• D. ${10}^{6}V$

Answer 1

The correct option is A $9V$

$K.E{.}_{max}=(h\gamma -\varphi )$

$\frac{1}{2}m{v}^2=h\gamma -\varphi$ ----------------(1)

$stoppingpotential=(h\gamma -\varphi )/e$
$V=(h\gamma -\varphi )/e$ ----------------(2)
Now, putting value of $(h\gamma -\varphi )$ in (2) from (1),

$V=\left(\frac{1}{2}m{v}^2\right)/e$

$=\frac{m{v}^2}{2e}$

$=\frac{v^2}{2e/m}$

$=\frac{1.8\times {10}^6\times \times 1.8\times {10}^6}{2\times 1.8\times {10}^{11}}$

$=9V.$
So, the answer is option (A).