Question

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

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

Hint:

Find the orbit number from the energy, then use Bohr's quantisation of angular momentum.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The energy of the nth orbit of hydrogen is \(E_n=-\dfrac{13.6}{n^{2}}\) eV. Angular momentum is quantised as \(L=\dfrac{nh}{2\pi}\).

Step 2: Find the Orbit
\(\dfrac{13.6}{n^{2}}=0.544\), so \(n^{2}=25\) and \(n=5\).

Step 3: Calculation
\(L=\dfrac{5h}{2\pi}\).

Final Answer:
The angular momentum is \(\dfrac{5h}{2\pi}\). \[ \boxed{\text{(B) }\dfrac{5h}{2\pi}} \]