Question

Electrons in a photoelectric-effect experiment emerge from an aluminum surface with a maximum kinetic energy of 1.30 eV. What is the wavelength of the light?

Short answer

Wavelength of the light is 222.35 nm

Step-by-step solution

Step 1. It is given that electrons emerge from an aluminum surface with a maximum kinetic energy of 1.30 eV 

We need to calculate the wavelength of light.

Step 2. Using the expression of maximum kinetic energy.

$$\begin{gathered} K_{max}=hf-E_0 \\ f=\frac{c}{\lambda } \\ {K}_{max}=h\frac{c}{\lambda }-E_0 \\ h\frac{c}{\lambda }=K_{max}+E_0 \\ \lambda =\frac{hc}{K_{max}+E_0} \\ \lambda =\frac{6.63\times {10}^{-34}\times 3\times {10}^{8}m/s}{1.3\times 1.6\times {10}^{-19}+4.28\times 1.6\times {10}^{-19}} \\ \lambda =222.35nm \end{gathered}$$