Question

In a photoelectric-effect experiment, the wavelength of light shining on an aluminum cathode is decreased from 250 nm to 200 nm. What is the change in the stopping potential?

Short answer

The change in stopping potential is 1.24 V

Step-by-step solution

Step 1. Given information is :Wavelength of light is decreased from 250 nm to 200 nm.

We need to find out the change in stopping potential.

Step 2. Using the expression for photoeffect.

$$\begin{gathered} \frac{hc}{\lambda }=E_k+E_0 \\ {E}_k=\frac{m{v}^2}{2}=e{V}_{stop} \\ \frac{hc}{\lambda }=e{V}_{stop}+E_0 \\ {V}_{stop}=\frac{\frac{hc}{\lambda }-E_0}{e} \\ {V}_{stop.i}=\frac{\frac{hc}{{\lambda }_i}-E_0}{e} \\ {V}_{stop.i}=\frac{\frac{\left(6.63\times {10}^{-34}\right)\left(3\times {10}^{8}m/s\right)}{250\times {10}^{-9}m}-\left(4.28\times 1.6\times {10}^{-19}\right)}{\left(1.6\times {10}^{-19}\right)} \\ {V}_{stop.i}=0.69V \\ {V}_{stop.f}=\frac{\frac{hc}{{\lambda }_f}-E_0}{e} \\ {V}_{stop.f}=\frac{\frac{\left(6.63\times {10}^{-34}\right)\left(3\times {10}^{8}m/s\right)}{200\times {10}^{-9}m}-\left(4.28\times 1.6\times {10}^{-19}\right)}{\left(1.6\times {10}^{-19}\right)} \\ {V}_{stop.f}=1.93V \\ ∆V_{stop}=V_{stop.f}-V_{stop.i} \\ ∆V_{stop}=1.93V-0.69V \\ ∆V_{stop}=1.24V \end{gathered}$$