Question 10

Light of frequency 7.21 × 10^{14} Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 10^{5} m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?

Answer

Frequency of the incident photon, v = 488 nm = 488 x 10^{-9} m

Maximum speed of the electrons, v = 6.0 × 10^{5} m/s

Planck’s constant, h = 6.626 × 10^{−34} Js

Mass of an electron, m = 9.1 × 10^{−31} kg

For threshold frequency ν0, the relation for kinetic energy is written as: *1/2 mv ^{2}*

= h(v - v_{0})

∴ v_{0} = v - mv^{2}/2h

= 7.21 x 10^{14} - (9.1 x 10^{-31}) x (6 x 10^{5})^{2} / 2 x (6.626 x 10^{-34})

= 7.21 x 10^{14} - 2.472 x 10^{14}

= 4.738 x 10^{14} Hz

*Therefore, the threshold frequency for the photoemission of electrons is 4.738 × 10 ^{14} Hz.*

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