If a hydrogen atom originally in a state with principal quantum number n is excited to state n' = 2n, then
A) its radius will quadruple and the binding energy will double.
B) its radius will quadruple and the binding energy will be reduced by a factor of four.
C) its radius and binding energy will quadruple.
D) its radius and binding energy will double.
E) its radius will double and the binding energy will quadruple.

Question

Quizplus · Accepted Answer

The radius of an electron orbit in a hydrogen atom is directly proportional to $n^2$, where $n$ is the principal quantum number. Therefore, if $n$ changes to $2n$, the radius becomes $(2n)^2 = 4n^2$, which means the radius quadruples. The binding energy, which is inversely proportional to $n^2$, will be reduced by a factor of four ($\frac{1}{(2n)^2} = \frac{1}{4n^2}$).

If a Hydrogen Atom Originally in a State with Principal