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: Use the ratio to the ground state:
$\frac{E_n}{E_1} = \frac{1}{n^2}$, so $\frac{0.544}{13.6} = \frac1{25}$.

Step 2: Conclude:
This gives $n = 5$.

Step 3: Quantisation:
The Bohr rule $mvr = \frac{nh}{2\pi}$ gives $\frac{5h}{2\pi}$.

Final Answer:
L equals 5h/(2 pi), option (C). \[ \boxed{\frac{5h}{2\pi}} \]