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_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) 800,000 m/s B) 212,000 m/s C) 653,000 m/s D) 591,000 m/s

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Quizplus · Accepted Answer

The maximum kinetic energy of the photoelectron is given by:$$KE_{max} = h
u - \phi$$where:* $$h$$ is Planck's constant* $$
u$$ is the frequency of the light* $$\phi$$ is the work functionThe frequency of the light is given by:$$
u = \frac{c}{\lambda}$$where:* $$c$$ is the speed of light* $$\lambda$$ is the wavelength of the lightSubstituting the given values into the equations, we get:$$KE_{max} = (6.63 	imes 10^{-34} J \cdot s)(3 	imes 10^8 m/s)/(200 	imes 10^{-9} m) - 5.22 eV$$$$KE_{max} = 9.94 	imes 10^{-19} J$$The maximum velocity of the photoelectron is given by:$$v = \sqrt{\frac{2KE_{max}}{m_e}}$$where:* $$m_e$$ is the mass of the electronSubstituting the given values into the equation, we get:$$v = \sqrt{\frac{2(9.94 	imes 10^{-19} J)}{9.11 	imes 10^{-31} kg}} = 5.91 	imes 10^5 m/s$$Therefore, the maximum velocity of the photoelectron is 591,000 m/s.

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