In the photoelectric effect, a photon with an energy of 5.3 × 10⁻¹⁹ J strikes an electron in a metal. Of this energy, 3.6 × 10⁻¹⁹ J is the minimum energy required for the electron to escape from the metal. The remaining energy appears as kinetic energy of the photoelectron. What is the velocity of the photoelectron, assuming it was initially at rest? A)3.7 × 10 ¹⁴ m/s B)3.7 × 10 ¹¹ m/s C)1.9 × 10 ⁶ m/s D)6.1 × 10 ⁵ m/s E)1.7 × 10 ⁻¹⁹ m/s

Question

Quizplus · Accepted Answer

The kinetic energy of the photoelectron can be found by subtracting the minimum energy required for the electron to escape from the total energy of the photon:
$KE = E_{photon} - E_{min} = (5.3 	imes 10^{-19} J) - (3.6 	imes 10^{-19} J) = 1.7 	imes 10^{-19} J$

We can then use the kinetic energy formula to find the velocity of the photoelectron:
$KE = \frac{1}{2}mv^2$
$1.7 	imes 10^{-19} J = \frac{1}{2}(9.11 	imes 10^{-31} kg)v^2$
$v = \sqrt{\frac{2(1.7 	imes 10^{-19} J)}{9.11 	imes 10^{-31} kg}}$
$v = 6.1 	imes 10^5 m/s = 6.1 	imes 10^{14} nm/s$

Therefore, the best choice is D.

In the Photoelectric Effect, a Photon with an Energy of 5.3