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: Understanding the Question:
We are given the total energy level of an electron in an excited state of a hydrogen atom ($E = -3.4\ \text{eV}$). We need to find the electron's orbital angular momentum ($L$) in this specific quantized state according to Bohr's model.

Step 2: Key Formula or Approach:
1. According to Bohr's energy quantization equation for hydrogen, the energy of the $n^{\text{th}}$ orbit is:
$$E_n = -\frac{13.6}{n^2}\ \text{eV}$$ 2. Bohr's second postulate states that orbital angular momentum is quantized as an integral multiple of $\frac{h}{2\pi}$:
$$L = \frac{nh}{2\pi}$$

Step 3: Detailed Explanation:
First, let's determine the principal quantum number $n$ using the given state energy value:
$$-3.4\ \text{eV} = -\frac{13.6}{n^2}\ \text{eV}$$ Isolate $n^2$ by rearranging the terms:
$$n^2 = \frac{-13.6}{-3.4} = 4$$ Taking the square root gives the principal quantum orbit level:
$$n = \sqrt{4} = 2$$ This tells us that the electron is orbiting in the second shell ($n=2$, which corresponds to the first excited state). Now, substitute $n = 2$ into Bohr's angular momentum quantization formula:
$$L = \frac{2 \cdot h}{2\pi}$$ The factor of 2 cancels out from the numerator and denominator, leaving:
$$L = \frac{h}{\pi}$$ This matches the expression provided in option (C).

Step 4: Final Answer:
The angular momentum of the electron in that excited state is $\frac{h}{\pi}$, corresponding to option (C).