Question

The ionization energy of the H atom is x KJ. The energy required for the electron to jump from n = 2 to n = 3 will be:
(A) $5x \ kJ$
(B) $\dfrac{36x}{5} \ kJ$
(C) $\dfrac{5x}{36} \ kJ$
(D) $\dfrac{9x}{4} \ kJ$

Answer 1

Hint: Electrons can jump from one energy level to another, but they can never have orbits with energies other than the allowed energy levels. Let's look at the simplest atom, a neutral hydrogen atom.

Complete step by step answer:
 The ionization energy of the H atom is x kJ.
Ionization energy is the quantity of energy that an isolated, gaseous atom in the ground electronic state must absorb to discharge an electron, resulting in a cation.
$H \to {{H}^ {+}} + {{e}^{-}}$
This energy is usually expressed in kJ/mol, or the amount of energy it takes for all the atoms in a mole to lose one electron each.
The energy of an electron in the nth orbit of hydrogen atom: $\dfrac {-13.6} {{{n}^ {2}}} eV$,
So, the ionization energy: +13.6eV= x kJ,
The energy required for the electron to jump from n=2 to n=3,
The energy difference of 2^nd and 3^rd orbit: $\Delta E=x\left [ \dfrac {1}{4}-\dfrac {1}{9} \right] kJ$,
Simplify the above relation:
$\Delta E=x\times \dfrac {5}{36} \ kJ$
So, the energy required for the electron to jump from n=2 to n=3 comes out to be: $\Delta E=\dfrac{5x}{36} kJ$,
Therefore, we can conclude that option (C) is correct.

Note: Excitation energy is the addition of a discrete amount of energy to a system such as an atomic nucleus, an atom, or a molecule that results in its alteration, ordinarily from the condition of lowest energy  to one of higher energy.