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?

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#### Solution

Frequency of the incident photon, `v = 488 nm = 488 xx 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 *ν*_{0}, the relation for kinetic energy is written as:

`1/2 mv^2 = h(v - v_0)`

`v_0 = v - (mv^2)/(2h)`

= `7.21 xx 10^14 - ((9.1 xx 10^(-31) )xx (6xx10^(5))^2)/(2xx(6.626 xx 10^(-34)))`

Therefore, the threshold frequency for the photoemission of electrons is `4.738 xx 10^14 Hz`

Concept: Photoelectric Effect and Wave Theory of Light

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