Question

If the frequency of light in a photoelectric experiment is doubles, the stopping potential with :

• A. Be doubled
• B. Halved
• C. Become more than doubled
• D. Become less than double

Answer 1

Answer: (C)

Become more than doubled

The energy equation is given as,

$e{V}_0=hv-W$

$V_0=\frac{hv}{e}-\frac{W}{e}$

When the frequency is doubled,

${V_0}^{\prime }=\frac{2hv}{e}-\frac{W}{e}$

${V_0}^{\prime }=2\left(\frac{hv}{e}-\frac{W}{e}\right)+\frac{W}{e}$

${V_0}^{\prime }=2V_0+\frac{W}{e}$

${V_0}^{\prime }>2V_0$

Thus, the stopping potential will become more than doubled.