Question

Please give step by step explanation because I have to learn how to solve similar problems on my own.

When radiation of 58.4nm from a helium discharge lamp is directed on a sample of krypton, electrons are ejected with a velocity of 1.59x10^6 m s^-1. The same radiation ejects electrons from rubidium atoms with a velocity of 2.45 x 10^6 m s^-1. What are the ionization energies (in eV of the two elements)?

Answer 1

By energy balance we have:

Ionization Energy (I.E) = Energy of incident radiation - Energy of electrons emitted

Energy of incident radiation (Er) = hc / λ

= (6.626 x 10¯³⁴ J s) (3.00 x 10⁸ m/s) / (58.4 x 10^-9 m) = 3.40 x 10^-18 J

= 3.40 x 10^-18 J * 6.242 x 10¹⁸ eV/J = 21.25 eV

Mass of an electron (Me) = 9.11 x 10^-31 kg

For Krypton:

Energy of electrons emitted = 0.5*Me*v²   (Kinetic energy)

= 0.5 * (9.11 x 10^-31 kg) (1.59 x 10⁶ m/s)² = 1.15 x 10^-18 J

= 1.15 x 10^-18 J * 6.242 x 10¹⁸ eV/J = 7.19 eV

I.E = 21.25 - 7.19 = 14.06 eV

For Rubidium:

Energy of electrons emitted = 0.5*Me*v²   (Kinetic energy)

= 0.5 * (9.11 x 10^-31 kg) (2.45 x 10⁶ m/s)² = 2.73 x 10^-18 J

= 2.73 x 10^-18 J * 6.242 x 10¹⁸ eV/J = 17.07 eV

I.E = 21.25 - 17.07 = 4.18 eV