Discussion
The ratio of the energy of the electron in the ground state of hydrogen atom to that of the electron in the first excited state of Be3+ is :
Answer: A
The ratio is 1 : 4.
For a one-electron atom or ion of nuclear charge Z, the energy of level n is
\[ E_{n} = -13.6\,\frac{Z^{2}}{n^{2}}\ \text{eV} \]
Hydrogen ground state: Z = 1, n = 1. Be3+ is one-electron beryllium, Z = 4, and its first excited state is n = 2:
\[ \begin{aligned} E_{H} &= -13.6\ \text{eV} \\ E_{Be^{3+}} &= -13.6 \times \tfrac{16}{4} = -54.4\ \text{eV} \end{aligned} \]
The ratio is 13.6 : 54.4 = 1 : 4. 1 : 16 forgets the n2 = 4 in the denominator (it is the ground-state ratio), and 1 : 8 and 2 : 9 come from using Z or n instead of their squares.
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