Chapter 27, Problem 44A

To determine

To Find: The maximum kinetic energy of an ejected photoelectron.

Answer to Problem 44A

  $1.06\times {10}^{-19}\text{ J}$

Explanation of Solution

Given data:

Wavelength of light, $\lambda \text{ =}6.5\times {10}^{\text{2}}\text{nm}$

Threshold frequency of the metal, $v_o=3.00\times {10}^{14}\text{Hz}$

Formula used:

Maximum kinetic energy,

  $K.E{.}_{\max }=h\upsilon -h{\upsilon }_0=\frac{hc}{\lambda }-h{\upsilon }_0$

Here, h is the Planck’s constant, c is the speed of light, $\lambda$ is the wavelength of incident light and $v_o$ is the threshold frequency.

Calculation:

  $\begin{array}{c}K.E_{\max }=\frac{hc}{\lambda }-h{v}_o\\ =\text{6}\text{.6}\times {\text{10}}^{-34}\left[\frac{3\times {10}^8}{6.5\times {10}^2\times {10}^{-9}}-3\times {10}^{14}\right]\\ =6.6\times {10}^{-34}\left[4.6\times {10}^{14}-3\times {10}^{14}\right]\\ =1.06\times {10}^{-19}\text{ J}\end{array}$

Conclusion:

Thus, the maximum kinetic energy of an ejected photoelectron is $1.06\times {10}^{-19}\text{ J}$ .