Question

If frequency of light in photoelectric experiment is tripled, the stopping potential will

• A. become more than triple
• B. become less than triple
• C. be tripled
• D. be $\frac{1}{3}rd$ of previous stopping potential

Answer 1

The correct option is A become more than triple

$K{E}_{max}=e{v}_{s_1}=h\nu -\varphi ...\left(1\right)$
$K{E}_{max2}=e{v}_{s_2}=3h\nu -\varphi$
$e{v}_{s_2}=3h\nu -\varphi$
$3e{v}_{s_1}=3h\nu -3\varphi$
$\frac{v_{s_2}}{3v_{s_1}}=\frac{3hv-\varphi }{3h\nu -3\varphi }>1$
$v_{s_2}>3v_{s_1}$