Question

The energy of an electron in the excited hydrogen atom is $-3.4\ \text{eV}$. Then according to Bohr's theory, the angular momentum of the electron in that excited state is ($h$ = Planck's constant)

• A. $\frac{2h}{\pi}$
• B. $\frac{nh}{2\pi}$
• C. $\frac{h}{\pi}$
• D. $\frac{3h}{2\pi}$

Hint:

Memorize the energy values of the first few shells of hydrogen to save time: $n=1 \rightarrow -13.6\ \text{eV}$, $n=2 \rightarrow -3.4\ \text{eV}$, $n=3 \rightarrow -1.51\ \text{eV}$. Recognizing $-3.4\ \text{eV}$ immediately tells you $n=2$, simplifying the angular momentum calculation to $\frac{2h}{2\pi} = \frac{h}{\pi}$ in just a few seconds.

Answer 1

The Correct Option is C

Step 1: Use Bohr's energy levels.
In a hydrogen atom the energy of the $n^{\text{th}}$ orbit is $$E_n = -\frac{13.6}{n^2}\ \text{eV}.$$
Step 2: Match the given energy.
We are told $E = -3.4\ \text{eV}$, so $$-3.4 = -\frac{13.6}{n^2}.$$
Step 3: Solve for $n^2$.
Cancelling the minus signs, $$n^2 = \frac{13.6}{3.4} = 4.$$
Step 4: Find the quantum number.
Taking the positive root, $n = 2$, the first excited state.
Step 5: Apply Bohr's angular-momentum rule.
Bohr's second postulate quantises angular momentum as $$L = \frac{n h}{2\pi}.$$
Step 6: Substitute $n = 2$.
$$L = \frac{2h}{2\pi} = \frac{h}{\pi}.$$ The factor of $2$ cancels neatly.
\[ \boxed{L = \frac{h}{\pi}} \]