Answer
The correct answer is:
(f) $~~~(a)~and~(b)$
Work Step by Step
We can write an expression for $K_{max}$, the maximum kinetic energy of an electron in the experiment:
$K_{max} = \frac{hc}{\lambda}-\phi$
The stopping potential is proportional to the value of $K_{max}$. Therefore, the stopping potential is determined by $\lambda$, the wavelength of the incident light, and $\phi$, the work function of the metal.
The correct answer is:
(f) $~~~(a)~and~(b)$
