Question

The angular momentum of an electron in Bohr's hydrogen atom having energy \((-0.544)\) eV is
(\(h\) = Planck's constant)

• A. \(\frac{h}{π}\)
• B. \(\frac{3h}{π}\)
• C. \(\frac{5h}{2π}\)
• D. \(\frac{7h}{2π}\)

Hint:

Find n from E = -13.6 / n squared, then L = n h / 2 pi.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In Bohr's hydrogen atom, the energy of level \(n\) is \(E_n = -\frac{13.6}{n^2}\) eV, and the angular momentum is \(L = \frac{nh}{2\pi}\).

Step 2: Find n:
\[ \frac{13.6}{n^2} = 0.544 \Rightarrow n^2 = \frac{13.6}{0.544} = 25 \Rightarrow n = 5 \]

Step 3: Angular momentum:
\[ L = \frac{nh}{2\pi} = \frac{5h}{2\pi} \]

Step 4: Why the other options are wrong.
\(\frac h\pi\) corresponds to \(n = 2\), \(\frac{3h}{\pi}\) to \(n = 6\), and \(\frac{7h}{2\pi}\) to \(n = 7\), each with a different energy from -0.544 eV.

Final Answer:
The angular momentum is \(\frac{5h}{2\pi}\), option (C). \[ \boxed{\frac{5h}{2\pi}} \]