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

| July 25, 2021 | education

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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