Determine the maximum speed of an electron emitted from potassium when illuminated by light

Determine the maximum speed of an electron emitted from potassium when illuminated by light with a wavelength of 400 nm, the work function of potassium is 2.2 eV.

λ = 400 nm = 400 * 10 ^ -9 m.
h = 6.6 * 10 ^ -34 J * s.
Avout = 2.2 eV = 3.52 * 10 ^ -19 J.
s = 3 * 10 ^ 8 m / s.
m = 9.1 * 10 ^ -31 kg.
V -?
The radiation energy E, which falls on potassium, goes to knock out electrons from the metal Avykh and give them kinetic energy Ek.
E = Avykh + Ek.
The energy of the incident radiation E is determined by the formula: E = h * c / λ, where h is Planck’s constant, c is the speed of light in vacuum, λ is the wavelength.
The kinetic energy of the outgoing electrons Ek is determined by the formula: Ek = m * V ^ 2/2, where m is the electron mass, V is the electron velocity.
h * c / λ = Avykh + m * V ^ 2/2.
m * V ^ 2/2 = h * s / λ – Avykh.
V = √ (2 * (h * s / λ – Avyh) / m).
V = √ (2 * (6.6 * 10 ^ -34 J * s * 3 * 10 ^ 8 m / s / 400 * 10 ^ -9 m – 3.52 * 10 ^ -19 J) / 9.1 * 10 ^ -31 kg) = 560,000 m / s.
Answer: the speed of the electron is V = 560,000 m / s.