You wish to pick an element for a photocell that will operate via the photoelectric effect with visible light. Which of the following are suitable (work functions are in parentheses): tantalum (4.2 eV), tungsten (4.5 eV), aluminium (4.2 eV), barium (2.5 eV), lithium (2.3 eV) ?

The given data is listed below.

The energy of different elements is given by,

Tantalum, 4.2 eV

Tungsten, 4.5 eV

Aluminium, 4.2 eV

Barium, 2.5 eV

Lithium,2.3 eV

Thewavelength of visible light is in the range of400 nm to 700 nm.

The wavelength of visible light is 400nm to 700nm.

Now, Energy is given by,

$E=\frac{hc}{\lambda }$ ….. (1)

Here,

The Plank’s constant is,

$$\begin{gathered} h=6.626\times {10}^{-34}{m}^2.kg/s \\ =\frac{6.626\times {10}^{-34}{m}^2.kg/s}{1.6\times {10}^{-19}J} \\ =4.141\times {10}^{-15}\mathrm{eV}.\mathrm{s} \end{gathered}$$

The speed of light is,

$$\begin{gathered} c=3\times {10}^{8}m/s \\ =3\times {10}^{17}nm/s \end{gathered}$$

Therefore, the value of hc is,

$$\begin{gathered} hc=4.14\times {10}^{-15}\mathrm{eV}.\mathrm{s}\times 3\times {10}^{17}\mathrm{nm}/\mathrm{s} \\ =12.40\times {10}^2\mathrm{eV}.\mathrm{nm} \\ =1240\mathrm{eV}·\mathrm{nm} \end{gathered}$$

Substitute 1240 eV. nm for hc and 400 nm for $\lambda$into equation (1), and you have

$\begin{array}{c}E=\frac{1240eV·nm}{400nm}\\ =3.1eV\end{array}$

Hence, Lithium and Barium are suitable to operate via photoelectric effect with visible light since their work functions are lower than 3.1 eV.