Question

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

• A. be doubled
• B. be halved
• C. become more than double
• D. become less than double

Answer 1

The correct option is C

become more than double

According to Einstein's photoelectric equation, $K.E_{max}$=hν−W, which tells us the value of the maximum kinetic energy corresponding to the fastest electron in the photocurrent. This fastest electron with the maximum kinetic energy gets just stopped when the potential is kept at the stopping potential $V_0$, i.e., the electron almost makes it to the collector plate before returning back to the emitter plate. Thus there is a point right near the collector plate where the whole of the kinetic energy, $K.E_{max}$ gets converted into potential energy due to the potential $V_0$. Thus, we can write -

$e{V}_0=K.E_{max}=h\nu -W$

$\Rightarrow v=\frac{e}{h}{V}_0+W$

Let $\nu \text{'}=2\nu$, the doubled frequency, for which the new stopping potential is, say, . We want to find the relationship between $V_0^{\prime }$ and $V_0$. Following the equation above, we can write , for $\nu \text{'}$ -

$v^{\prime }=\frac{e}{h}{V}_0^{\prime }+W$ .....(1)

But also,

$v^{\prime }=2v=\frac{e}{h}\left(2V_0\right)+2W=\frac{e}{h}(2V_0+\frac{hW}{e})+W$ ....(2)

Comparing the equations (1) and (2),

$\Rightarrow V_0^{\prime }=2V_0+\frac{hW}{e}>2V_0$

As you can see, option (c) is the correct choice!