Question

If the ratio of the time periods of the electrons revolving in the first and \(n^{\text{th}}\) orbits of Hydrogen atom is \(1:64\), then the angular momentum of the electron in the \(n^{\text{th}}\) excited state of Hydrogen atom is \((h=\text{Planck's constant})\)

• A. \(\dfrac{3.5h}{\pi}\)
• B. \(\dfrac{5h}{\pi}\)
• C. \(\dfrac{2.5h}{\pi}\)
• D. \(\dfrac{2h}{\pi}\) \bigskip

Hint:

For Hydrogen atom: \[ T_n\propto n^3, \] \[ r_n\propto n^2, \] \[ v_n\propto \frac{1}{n}, \] and \[ L_n=\frac{nh}{2\pi}. \] Remember that the \(k^{\text{th}}\) excited state corresponds to orbit number \(k+1\).

Answer 1

The Correct Option is C

Concept: Bohr model: \(T_n \propto n^3\). Find \(n\) from period ratio, then angular momentum \(L = nh/2\pi\).

Step 1: \(T_n/T_1 = n^3 = 64 \Rightarrow n=4\). The \(n\)th excited state corresponds to orbit \(n+1 = 5\). \(L = 5h/2\pi\).

Step 2: Write the final answer. \(\boxed{L=\frac{5h}{2\pi}}\)