Question

The angular momentum of the electron in the third Bohr orbit of a hydrogen atom is '\(l\)' so its angular momentum in the fourth Bohr orbit is

• A. \(4l\)
• B. \((\frac{5}{4})l\)
• C. \((\frac{4}{3})l\)
• D. \((\frac{3}{2})l\)

Hint:

Angular momentum is n h over 2 pi, so it is proportional to n.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Bohr quantization: \(L = \frac{nh}{2\pi}\), so \(L \propto n\).

Step 2: Key Formula or Approach:
\(L_3 = l\) for \(n = 3\). Find \(L_4\) for \(n = 4\).

Step 3: Detailed Explanation:
\[ \frac{L_4}{L_3} = \frac43 \Rightarrow L_4 = \frac43\,l \]
Option B, \(\frac54 l\), would be wrong; the ratio is of the orbit numbers \(4\) and \(3\).
The orbit number \(n\) is the only quantity that changes, since \(h\) and \(2\pi\) are constants. So angular momentum grows in the same proportion as \(n\), going from \(3\) units to \(4\) units, a ratio of \(\frac43\).

Final Answer:
The angular momentum is \(\frac{4}{3}\,l\), option (C). \[ \boxed{\frac{4}{3}\,l} \]