Answer to Question #240223 in Electricity and Magnetism for yuyun

Question

A ray of purple light hits the surface of a metal which has energy
threshold (2/3) times the energy quanta of violet light. If the frequency of the    purple light is 10^16 Hz and Plank's constant = 6.626.10^−34 , determine the kinetic energy of electrons escaping from the metal surface.

Answer 1

Since we know that the energy of the violet light is 2/3 the threshold, then W=(2/3)$E_{\text{violet light}}$ and we can calculate the kinetic energy with:

$E_{\text{photoelectric effect}}=E_{\text{violet light}}=W+E_k \\ \implies E_k=E_{\text{photoelectric effect}}-W=E_{\text{violet light}}-(2/3)E_{\text{violet light}} \\ \implies E_k=E_{\text{violet light}}/3=\cfrac{h\nu}{3}$

We substitute and find E_k:

$E_k= \cfrac{(6.626\times10^{-34}Js)(10^{16}\,s^{-1})}{3}=2.20866 \times10^{-18}J$

In conclusion, the kinetic energy of the electrons escaping from the metal surface is E_K = 2.209 X 10^-18 J.

Reference

• Chang, R., & Goldsby, K. A. (2010). Chemistry. Chemistry, 10th ed.; McGraw-Hill Education: New York, NY, USA.