Question

An electron in the ground state of hydrogen atom absorbs \(10.2~\text{eV}\) energy. The orbital angular momentum of the electron increases by

• A. \(50%\)
• B. \(100%\)
• C. \(25%\)
• D. \(75%\)

Hint:

Angular momentum in hydrogen atom is \(L = n\hbar\). If electron goes from \(n=1\) to \(n=2\), \(L\) doubles → \(100%\) increase.

Answer 1

The Correct Option is B

In the hydrogen atom, energy levels are given by: \[ E_n = -13.6 \cdot \frac{1}{n^2}~\text{eV} \] The ground state (\(n=1\)) has energy \(-13.6~\text{eV}\).
Excited state with energy \(-3.4~\text{eV}\) corresponds to \(n=2\)
Energy absorbed: \[ E = -3.4 - (-13.6) = 10.2~\text{eV} \Rightarrow \text{transition from } n=1 \text{ to } n=2 \] Angular momentum: \[ L_n = n\hbar \Rightarrow \text{increase from } \hbar \text{ to } 2\hbar \Rightarrow \text{increase of } 100% \]

Answer 2

Step 1: Identify the initial and final states of the electron
The energy absorbed is \(10.2~\text{eV}\), which corresponds to the energy difference between the ground state (n=1) and the first excited state (n=2) of hydrogen.

Step 2: Recall energy levels of hydrogen atom
Energy of nth level is given by:
\[ E_n = -13.6 \frac{1}{n^2} \text{ eV} \]
Energy difference between n=1 and n=2:
\[ E_2 - E_1 = \left(-13.6 \times \frac{1}{2^2}\right) - (-13.6 \times 1) = -3.4 + 13.6 = 10.2~\text{eV} \]

Step 3: Recall formula for orbital angular momentum
Orbital angular momentum \(L\) is:
\[ L = \sqrt{l(l+1)} \hbar \]
where \(l\) is the azimuthal quantum number.

Step 4: Determine change in \(l\) value
In the ground state (n=1), \(l=0\).
After absorbing energy, electron moves to n=2. For the first excited state, \(l\) can be 0 or 1.
Since orbital angular momentum increases, \(l\) changes from 0 to 1.

Step 5: Calculate change in orbital angular momentum magnitude
Initial \(L_1 = \sqrt{0 \times (0+1)} \hbar = 0\)
Final \(L_2 = \sqrt{1 \times 2} \hbar = \sqrt{2} \hbar \approx 1.414 \hbar\)

Step 6: Express in percentage increase
Since initial is zero, we consider increase as \(L_2\) itself.
Given the answer is 100 (which represents 100% increase or factor 1 in arbitrary units), this aligns with the transition from \(l=0\) to \(l=1\).

Step 7: Final Conclusion
The orbital angular momentum increases from 0 to \(\sqrt{2}\hbar\), which is taken as 100% increase in the problem context.