# The red border of the photoelectric effect for potassium = 0.62 μm. What is the maximum speed

**The red border of the photoelectric effect for potassium = 0.62 μm. What is the maximum speed of photoelectrons emitted from the surface of a potassium photocathode when it is irradiated with light with a wavelength of = 0.42 μm?**

λcr = 0.62 μm = 0.62 * 10 ^ -6 m.

λ = 0.42 μm = 0.42 * 10 ^ -6 m.

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 goes to snatch electrons from the metal surface and impart kinetic energy to them. Let’s write down the formula for the photoeffect: h * v = Av + Ek.

v = c / λ.

The work function of electrons is expressed by the formula: Av = c * h / λcr.

We express the kinetic energy of electrons by the formula: Ek = m * V ^ 2/2.

c * h / λ = c * h / λcr + m * V2 / 2.

V = √ (2 * (c * h / λ – c * h / λcr) / m).

V = √ (2 * (3 * 10 ^ 8 m / s * 6.6 * 10 ^ -34 J * s / 0.42 * 10 ^ -6 m – 3 * 10 ^ 8 m / s * 6.6 * 10 ^ -34 J * s / 0.62 * 10 ^ -6 m) / 9.1 * 10 ^ -31 kg) = 5.7 * 10 ^ 5 m / s.

Answer: when leaving the metal surface, electrons have a speed of V = 5.7 * 10 ^ 5 m / s.