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Question
Light of frequency 7.21 × 1014 Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 105 m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?
Solution
Frequency of the incident photon, v = 488 nm = 488 × 10−9 m
Maximum speed of the electrons, v = 6.0 × 105 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)/(2"h")`
= `7.21 xx 10^14 - ((9.1 xx 10^(-31) ) xx (6 xx 10^(5))^2)/(2 xx (6.626 xx 10^(-34)))`
= 7.21 × 1014 − 2.472 × 1014
= 4.738 × 1014 Hz
Therefore, the threshold frequency for the photoemission of electrons is 4.738 × 1014 Hz.
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