What is the maximum velocity of a photoelectron emitted from a surface with work function 3.5 eV when illuminated by 289 nm ultraviolet light? (m_electron = 9.11 × 10−³¹ kg,h = 6.63 × 10−³⁴ J⋅s,1 eV = 1.60 × 10−¹⁹ J,and 1 nm = 10−⁹ m) A)3.75E+5 m/s B)5.31E+5 m/s C)7.28E+2 m/s D)8.62E+5 m/s E)1.60E+6 m/s

Question

Quizplus · Accepted Answer

First, we need to find the energy of the incoming photon:
E = hc/λ = (6.63x10^-34 J⋅s)(3x10^8 m/s)/(289x10^-9 m) = 2.16x10^-18 J

Next, we can use the work function and the photon energy to find the maximum kinetic energy of the emitted electron:
KEmax = E - work function = 2.16x10^-18 J - (3.5 eV)(1.60x10^-19 J/eV) = 1.476x10^-18 J
(Note that we converted the work function from eV to J)

Finally, we can use the maximum kinetic energy to find the maximum velocity of the photoelectron:
KEmax = 1/2 mv^2
v = sqrt(2KEmax/m) = sqrt((2)(1.476x10^-18 J)/(9.11x10^-31 kg)) = 5.31x10^5 m/s

Therefore, the correct answer is B.

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