Question

It takes 208.4 kJ of energy to remove 1 mole of electrons from the atoms on the surface of rubidium metal. If rubidium metal is irradiated with 254-nm light, what is the maximum kinetic energy the released electrons can have?

Answer 1

Answer:

4.34x10⁻¹⁹ J

Explanation:

The total energy emitted by irradiation is given by

E = hf, where E is the energy, h is the plack constant (6.626x10⁻³⁴ J.s), and f is the frequency. The frequency is also the velocity of the light (c = 2.99x10⁸ m/s) divided by the length of the irradiation (254x10⁻⁹ m). So:

E = (6.626x10⁻³⁴)x(2.99x10⁸)/ (254x10⁻⁹)

E = 7.80x10⁻¹⁹ J

The energy to remove 1 electron is the energy necessary to remove 1 mol divided by the Avogadros number ( 1 mol = 6.02x10²³ electrons):

208400/6.022x10^23 = 3.46x10⁻¹⁹ J

The total energy is the energy necessary to remove one electrons plus the kinectic energy (Ek) of the electrons:

7.80x10⁻¹⁹ = 3.46x10⁻¹⁹ + Ek

Ek = 7.80x10⁻¹⁹ - 3.46x10⁻¹⁹

Ek = 4.34x10⁻¹⁹ J