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 × 10¹⁴ m/s B) 3.7 × 10¹¹ m/s C) 1.9 × 10⁶ m/s D) 6.1 × 10⁵ m/s E) 1.7 × 10^-19 m/s

Question

Quizplus · Accepted Answer

Using conservation of energy:

Energy of photon = Work function + Kinetic energy of photoelectron

5.3 × 10^-19 J = 3.6 × 10^-19 J + (1/2)mv^2

Solving for v:

v = sqrt[(2(5.3 × 10^-19 J - 3.6 × 10^-19 J)/m)]

The mass of an electron is 9.11 × 10^-31 kg. Plugging this in:

v = 6.1 × 10^5 m/s

Rounding to two significant figures, the answer is 6.1 × 10^5 m/s or 6.1 × 10^15 m/s.

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