# Determine the speed of the electron if the work function is 2 eV, it is irradiated with light with a wavelength of 300nM.

m = 9.1 * 10 ^ -31 kg.

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

Av = 2 eV = 3.2 * 10 ^ -19 J.

λ = 300 nm = 300 * 10 ^ -9 m.

c = 3 * 10 ^ 8 m / s.

V -?

Let’s write down the law of the photoelectric effect: h * v = Av + Ek.

The frequency of the incident photons v is expressed by the formula: v = c / λ, where c is the speed of light, λ is the wavelength of the photons.

We express the kinetic energy of electrons Ek, which fly out of the metal surface, by the formula: Ek = m * V ^ 2/2.

h * c / λ = Ab + m * V ^ 2/2.

m * V ^ 2/2 = h * c / λ – Av.

V = √ (2 * (h * c / λ – Av) / m).

V = √ (2 * (6.6 * 10 ^ -34 J * s * 3 * 10 ^ 8 m / s / 300 * 10 ^ -9 m – 3.2 * 10 ^ -19 J) / 9.1 * 10 ^ -31 kg) = 1.2 * 10 ^ 6 m / s.

Answer: electrons fly out of the metal surface at a speed of V = 1.2 * 10 ^ 6 m / s.

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