Question

Relation between the stopping potential $V_o$ of a metal and the maximum velocity ${}^{\prime }v{}^{\prime }$ of the photoelectrons is

• A. $V_0\propto \frac{1}{v^2}$
• B. $V_0\propto v^2$
• C. $V_0\propto v$
• D. $V_0\propto \frac{1}{v}$

Answer 1

The correct option is B $V_0\propto v^2$

$K.E{.}_{max}=h\gamma -\varphi$
$\frac{1}{2}m{v}^2=h\gamma -\varphi$ ------------(1)

$stoppingpotential=(h\gamma -\varphi )/e$

$V_o=(h\gamma -\varphi )/e$ ------------(2)
From (1) put value of $(h\gamma -\varphi )$ in (2),

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

$V_o=\frac{m{v}^2}{2e}$
So, $V\propto v^2$
So, the answer is option (B).