Question

When (5.84 nm) radiation from a helium discharge lamp is directed on a sample of krypton, electrons are ejected with a velocity of (1.59 m/s). The same radiation ejects electrons from rubidium atoms with a velocity of (2.45 m/s). What are the ionization energies of krypton and rubidium?

a) Ionization energy of krypton = (5.84 nm), ionization energy of rubidium = (2.45 m/s)

b) Ionization energy of krypton = (2.45 m/s), ionization energy of rubidium = (5.84 nm)

c) Ionization energy of krypton = (1.59 m/s), ionization energy of rubidium = (2.45 m/s)

d) Ionization energy of krypton = (2.45 m/s), ionization energy of rubidium = (1.59 m/s)

Answer 1

The ionization energy of krypton and rubidium can be calculated using the velocity of the ejected electrons when exposed to radiation from a helium discharge lamp.

Step-by-step explanation:

The ionization energy of an atom refers to the amount of energy required to remove an electron from the atom. In this question, we are given that electrons are ejected from krypton and rubidium atoms when exposed to 5.84 nm radiation from a helium discharge lamp. The velocity of the ejected electrons can be used to determine the ionization energy of each atom.

The ionization energy of krypton can be calculated by using the formula:

Ionization Energy = 1/2 * mass * (velocity)^2

Substituting the given values, Ionization Energy of krypton = 1/2 * mass * (1.59 m/s)^2

Similarly, the ionization energy of rubidium can be calculated using the velocity of the ejected electrons from rubidium atoms.

Ionization Energy of rubidium = 1/2 * mass * (2.45 m/s)^2

Therefore, the correct answer is d) Ionization energy of krypton = (2.45 m/s), ionization energy of rubidium = (1.59 m/s).