Chapter 27, Problem 20P

To determine

The metal which will not emit electrons, when visible light shines on it?

Answer to Problem 20P

Solution:

The metals which will not emit electrons are copper and iron.:

Explanation of Solution

Given:

Work function of sodium, $W_{\text{Na}}=2.3\text{eV}$

Work function of cesium, $W_{\text{CS}}=2.1\text{eV}$

Work function of copper, $W_{\text{Cu}}=4.7\text{eV}$

Work function of iron, $W_{\text{Fe}}=4.5\text{eV}$

Wavelength range of visible spectrum, $\lambda =400\sim 750\text{nm}$

Formula used:

The work function is calculated by the formula

$W=\frac{hc}{e\lambda }\left(\text{eV}\right)$

Here,

$e$ is the charge of electron

$h=6.62\times {10}^{-34}\text{J-s}$ is the Planck’s constant

$c=3\times {10}^8\text{m/s}$ is the speed of light

$\lambda$ is the wavelength

Calculation:

The wave function is defined as the minimum energy required by the photon to emit electron from the metal surface.

The maximum energy that visible electron can supply to any metal is calculated as

$W=\frac{hc}{e\lambda }\left(\text{eV}\right)$

Plugging the values in the above equation

$\begin{array}{c}W=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{1.6\times {10}^{-19}\times 400\times {10}^{-9}}\\ =3.10\text{eV}\end{array}$

The metal whose work function is greater than the energy supplied by the visible spectrum will not emit electron.

Since it’s clear that

$W_{\text{Cu}},W_{\text{Fe}}>W$

Thus, copper and iron will not emit electron when light of visible spectrum strikes on the metal.