# Determine the maximum rate of emission of photoelectrons from potassium, the work function of electrons

Determine the maximum rate of emission of photoelectrons from potassium, the work function of electrons of which is 3.6 * 10 ^ (- 19) J when illuminated with ultraviolet light with a wavelength of 200nm.

λ = 200 nm = 200 * 10 ^ -9 m.

Av = 3.6 * 10 ^ -19 J.

h = 6.6 * 10 ^ -34 J * s.

m = 9.1 * 10 ^ -31 kg.

C = 3 * 10 ^ 8 m / s.

V -?

The energy of the incident photons Eph goes to knock out electrons from the surface of the metal Av and impart kinetic energy Ek to them.

Eph = Av + ​​Ek – the law of the photoelectric effect.

We express the kinetic energy of the emitted electrons Ek by the formula: Ek = m * V ^ 2/2, where m is the electron mass, V is the rate of electron ejection from the metal surface.

The energy of photons Eph is expressed by the formula: Eph = h * C / λ, where h is Planck’s constant, C is the speed of light, λ is the wavelength of photons.

h * C / λ = Ab + m * V ^ 2/2.

m * V ^ 2/2 = h * C / λ – Av.

The electron emission speed V is expressed by the formula: V = √ (2 * h * C / λ * m – 2 * Av / m).

V = √ (2 * 6.6 * 10 ^ -34 J * s * 3 * 10 ^ 8 m / s / 200 * 10 ^ -9 m * 9.1 * 10 ^ -31 kg – 2 * 3.6 * 10 ^ -19 J / 9.1 * 10 ^ -31 kg) = 1.14 * 10 ^ 6 m / s.

Answer: the maximum speed of electron emission is V = 1.14 * 10 ^ 6 m / s.

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