Chapter 27, Problem 48A

To determine

To Find: Work function of the cathode.

Answer to Problem 48A

  $2.9\times {10}^{-19}\text{J}$

Explanation of Solution

Given data:

Wavelength of red light, $\lambda =680\text{nm}$

Formula used:

Work function of the metal, $\varphi =\frac{hc}{{\lambda }_0}$

Here, h is the Planck’s constant, c is the speed of light and ${\lambda }_o$ is the threshold frequency.

Calculation:

Red color light has the lowest frequency in the visible range.

The work function $\varphi =\frac{hc}{{\lambda }_0}$

  $\begin{array}{l}\varphi =\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{680\times {10}^{-9}}\\ \varphi =0.029\times {10}^{-34+17}\\ \varphi =0.029\times {10}^{-17}\text{J}\\ \varphi =2.9\times {10}^{-19}\text{J}\end{array}$

Conclusion:

The work function of the cathode is $\varphi =2.9\times {10}^{-19}\text{J}$ .