What is the kinetic energy and speed of a photoelectron emitted from magnesium

What is the kinetic energy and speed of a photoelectron emitted from magnesium (Av = 3.46 eV) when it is irradiated with ultraviolet light with a wavelength of 340 nm?

Given:

me = 9.1 * 10 ^ -31 kg;

Avih = 3.46 eV = 3.46 * 1.6 * 10 ^ -19 J = 5.5 * 10 ^ -19 J;

λ = 340 nm = 340 * 10 ^ -9 m = 3.4 * 10 ^ -7 m.

To find:

Wk -?

V -?

Decision.

Find the speed using the following formula:

hν = Abih + me * V ^ 2/2;

6.62 * 10 ^ -34 * v = 5.5 * 10 ^ -19 + 9.1 * 10 ^ -31 * V ^ 2/2;

6.62 * 10 ^ -34 * c / λ = 5.5 * 10 ^ -19 + 9.1 * 10 ^ -31 * V ^ 2/2;

6.62 * 10 ^ -34 * 3 * 10 ^ 8 / 3.4 * 10 ^ -7 = 5.5 * 10 ^ -19 + 9.1 * 10 ^ -31 * V ^ 2/2;

19.86 * 10 ^ -26 / 3.4 * 10 ^ -7 = 5.5 * 10 ^ -19 + 9.1 * 10 ^ -31 * V ^ 2/2;

5.8 * 10 ^ -19 = 5.5 * 10 ^ -19 + 9.1 * 10 ^ -31 * V ^ 2/2;

5.8 * 10 ^ -19 – 5.5 * 10 ^ -19 = 9.1 * 10 ^ -31 * V ^ 2/2;

0.3 * 10 ^ -19 = 9.1 * 10 ^ -31 * V ^ 2/2;

0.3 * 10 ^ -19 * 2 = 9.1 * 10 ^ -31 * V ^ 2;

0.6 * 10 ^ -19 = 9.1 * 10 ^ -31 * V ^ 2;

V2 = 0.6 * 10 ^ -19: 9.1 * 10 ^ -31;

V2 = 0.07 * 10 ^ 12;

V = 0.26 * 10 ^ 6 m / s.

Let’s find the kinetic energy:

Wk = me * V ^ 2/2 = 9.1 * 10 ^ -31 * 0.07 * 10 ^ 12/2 = 0.32 * 10 ^ -19 J.