Answer
Cesium
Work Step by Step
We can find the required suitable material as follows:
The threshold wavelength $(\lambda_{\circ})$ for aluminum is given as
$\lambda_{\circ}=\frac{hc}{W_{\circ}}$
$\implies \lambda=\frac{(6.63\times 10^{-34}J.s)(3.0\times 10^8m/s)}{4.28eV}$
$\implies \lambda=290.4nm$
This wavelength is not in the visible light wavelength range.
Now for Lead
$\lambda_{\circ}=\frac{hc}{W_{\circ}}$
$\implies \lambda_{\circ}=\frac{(6.63\times 10^{-34}J.s)(3.0\times 10^8m/s)}{4.25eV}$
$\implies \lambda_{\circ}=292.5nm$
This wavelength is not in the visible wavelength range.
For Cesium, the threshold wavelength is given as
$\lambda_{\circ}=\frac{hc}{W_{\circ}}$
We plug in the known values to obtain:
$\lambda_{\circ}=\frac{(6.63\times 10^{-34}J.s)(3.0\times 10^8m/s)}{(2.14eV)(1.6\times 10^{-19}J/eV)}$
$\implies \lambda_{\circ}=580.9nm$
This wavelength is in the visible light wavelength range. Thus, Cesium metal surface is suitable for ejecting electrons from the metal surface for visible light.
