When the frequency of incident light in a photoelectric experiment...

Einstein's photoelectric equation is given by $eV_{s} = h\nu - \Phi$, where $V_{s}$ is the stopping potential, $\nu$ is the frequency of incident light, and $\Phi$ is the work function of the material.

For the initial frequency $\nu$, the stopping potential $V_{s1}$ is:

$V_{s1} = \dfrac{h\nu}{e} - \dfrac{\Phi}{e}$

When the frequency is doubled to $2\nu$, the new stopping potential $V_{s2}$ is:

$V_{s2} = \dfrac{h(2\nu)}{e} - \dfrac{\Phi}{e}$

Rewriting the above equation:

$V_{s2} = \dfrac{2h\nu}{e} - \dfrac{2\Phi}{e} + \dfrac{\Phi}{e}$

$V_{s2} = 2\left(\dfrac{h\nu}{e} - \dfrac{\Phi}{e}\right) + \dfrac{\Phi}{e}$

$V_{s2} = 2V_{s1} + \dfrac{\Phi}{e}$

Since the work function $\Phi$ is a positive quantity, $\dfrac{\Phi}{e} > 0$.

Therefore, $V_{s2} > 2V_{s1}$, which implies that the stopping potential becomes more than double.