Question

Which metals in Table 38.1 exhibit the photoelectric effect for (a) light with$\mathrm{\lambda }=400\mathrm{nm}$ and (b) light with $\mathrm{\lambda }=250\mathrm{nm}$?

Short answer

a. Sodium

b. Iron.

Step-by-step solution

Part (a) Step 1: Given information

We have given,

$\mathrm{\lambda }=400\mathrm{nm}$

We have to find the metal for the photo electric effect.

Simplify

For $\mathrm{\lambda }=400\mathrm{nm}$, the energy will be

$$\begin{gathered} \mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }} \\ \mathrm{E}=\frac{1240\mathrm{eV}.\mathrm{nm}}{400\mathrm{nm}} \\ \mathrm{E}=3.1\mathrm{eV} \end{gathered}$$

Then we can say that the sodium will be the metal for which we need the$400\mathrm{nm}$wavelength source light.

Part (b) Step 1: Given information

We have given,

$\mathrm{\lambda }=250\mathrm{nm}$

We have to find the metal for the photoelectric effect.

Simplify

The energy will be given by,

$$\begin{gathered} \mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }} \\ \mathrm{E}=\frac{1240\mathrm{eV}.\mathrm{nm}}{250\mathrm{nm}} \\ \mathrm{E}=4.96\mathrm{eV} \end{gathered}$$

For this much energy we need the iron metal.