Question

A graph is drawn between frequency of the incident radiation (on X- axis) and stopping potential (on Y-axis). Then the slope of the straight line indicates

• A. $h.e$
• B. $h/e$
• C. $e/h$
• D. $(e-h)$

Answer 1

Answer: (B)

$h/e$

$hv=h{v}_0+K.E$
but at stopping potential $K.E=eV$
$\Rightarrow hv=h{v}_0+eV$
$\Rightarrow V=\frac{h}{e}(v-v_0)$
$\therefore$ slope is $\frac{h}{e}$