Question

The energy of an electron in an orbit in hydrogen atom is -3.4 eV. Its angular momentum in the orbit will be:

• A. \(\frac{3h}{2\pi}\)
• B. \(\frac{2h}{\pi}\)
• C. \(\frac{h}{\pi}\)
• D. \(\frac{h}{2\pi}\)

Hint:

Memorizing the energy levels for Hydrogen helps save time: $n=1$ is $-13.6$ eV, $n=2$ is $-3.4$ eV, and $n=3$ is $-1.51$ eV.

Answer 1

The Correct Option is C

To solve this problem, we need to determine the angular momentum of an electron in a specific orbit of a hydrogen atom, given its energy. The energy of an electron in an orbit of hydrogen is -3.4 eV.

The energy levels of an electron in the hydrogen atom are quantized and can be expressed by the formula:

\(E_n = -\frac{13.6 \, \text{eV}}{n^2}\)

where \(n\) is the principal quantum number.

Given the energy \(E_n = -3.4 \, \text{eV}\), we equate and solve for \(n\):

\(-3.4 = -\frac{13.6}{n^2}\)

Solving for \(n^2\):

\(3.4 = \frac{13.6}{n^2} \\ n^2 = \frac{13.6}{3.4} \\ n^2 = 4 \\ n = 2\)

So, the principal quantum number \(n\) is 2.

The angular momentum for an electron in a quantized orbit is given by:

\(L = n\frac{h}{2\pi}\)

Substituting \(n = 2\) gives:

\(L = 2 \cdot \frac{h}{2\pi} = \frac{2h}{2\pi} = \frac{h}{\pi}\)

Therefore, the angular momentum of the electron in this orbit is \(\frac{h}{\pi}\), which matches the correct answer.

Thus, the correct option is:

\(\frac{h}{\pi}\)