What is the maximum velocity of a photoelectron emitted from a surface with work function 5.22 eV when illuminated by 200-nm ultraviolet light? (m_electro_n = 9.11 *10^-³¹ kg, h = 6.63 *10^-³⁴ J.s, 1 eV = 1.60 *10^-¹⁹J, and 1 nm = 10^-⁹m) A) 800,000 m/s B) 212,000 m/s C) 653,000 m/s D) 591,000 m/s

Question

Quizplus · Accepted Answer

The maximum kinetic energy of the photoelectron is given by:KE = hc/λ - Φwhere h is Planck's constant, c is the speed of light, λ is the wavelength of the incident light, and Φ is the work function of the surface.Plugging in the given values, we get:KE = (6.63 × 10^-34 J s)(3 × 10^8 m/s)/(200 × 10^-9 m) - 5.22 eV(1.60 × 10^-19 J/eV)= 9.94 × 10^-19 JThe maximum velocity of the photoelectron is then:v = √(2KE/m)= √(2(9.94 × 10^-19 J)/(9.11 × 10^-31 kg))= 5.91 × 10^5 m/sTherefore, the correct answer is D) 591,000 m/s.

What Is the Maximum Velocity of a Photoelectron Emitted from a Surface