Question

The radii of two of the first four Bohr's orbits of the hydrogen atom are in the ratio 1 : 4. The energy difference between them may be:

• A. Either 12.09 eV or 10.2 eV
• B. Either 2.55 eV or 10.2 eV
• C. Either 13.6 eV or 3.4 eV
• D. Either 3.4 eV or 0.85 eV

Answer 1

The correct option is B Either 2.55 eV or 10.2 eV

$\frac{R_{n_1}}{R_{n_2}}=\frac{n_1^2}{n_2^2}=\frac{1}{4},\therefore \frac{n_1}{n_2}=\frac{1}{2},\frac{2}{4},\frac{3}{6}$ etc.
Among the first four orbits $n_1$ and $n_2$ can be 1 and 2 or 2 and 4.
$\therefore$ Energy difference can be:
$E_{2\to 1}=10.2eV$ or $E_{4\to 2}=2.55eV$