Chapter 27, Problem 65A

To determine

To Calculate: Maximum kinetic energy of the photoelectron.

Answer to Problem 65A

  $\text{52}\text{.96}\times {\text{10}}^{-20}\text{J}$

Explanation of Solution

Given:

Threshold frequency of the metal, $f=8.0\times {10}^{14}\text{Hz}$

Frequency of the light, $f_l=1.6\times {10}^{15}\text{Hz}$

Formula Used:

Maximum kinetic energy of the photoelectrons is:

  $K.E_{\max }=h{f}_l-hf$

Calculations:

Substitute the values and solve.

  $\begin{array}{l}K.E_{\max }=h\left(f_l-f\right)\\ K.E_{\max }=6.62\times {10}^{-34}\text{J}\cdot \text{s}\left(\text{1}\text{.6}\times {\text{10}}^{15}\text{Hz -8}\text{.0}\times {\text{10}}^{14}\text{Hz}\right)\\ K.E_{\max }=\text{52}\text{.96}\times {\text{10}}^{-20}\text{J}\end{array}$

Conclusion:

Thus, the maximum kinetic energy of the photoelectrons is $\text{52}\text{.96}\times {\text{10}}^{-20}\text{J}$ .