In an `e^(-)` transition inside a hydrogen atom, orbital angular momentum may change by (h=Planck's constant)

Options

A. h
B. `(h)/(pi)`
C. `(h)/(2pi)`
D. `(h)/(4pi)`

Answer

To solve the problem regarding the change in orbital angular momentum during an electron transition inside a hydrogen atom, we will follow these steps:

### Step 1: Understand the Angular Momentum in Bohr's Model
According to Bohr's model of the hydrogen atom, the angular momentum (L) of an electron in a given orbit is quantized and given by the formula:
\[ L = n \frac{h}{2\pi} \]
where \( n \) is the principal quantum number (1, 2, 3, ...), and \( h \) is Planck's constant.

### Step 2: Calculate Angular Momentum for Different States
For an electron in the nth orbit, the angular momentum can be expressed as:
- For \( n_1 \):
\[ L_1 = n_1 \frac{h}{2\pi} \]
- For \( n_2 \):
\[ L_2 = n_2 \frac{h}{2\pi} \]

### Step 3: Determine the Change in Angular Momentum
The change in angular momentum (\( \Delta L \)) during a transition from one state to another can be calculated as:
\[ \Delta L = L_2 - L_1 \]
Substituting the expressions for \( L_1 \) and \( L_2 \):
\[ \Delta L = \left(n_2 \frac{h}{2\pi}\right) - \left(n_1 \frac{h}{2\pi}\right) \]
\[ \Delta L = \frac{h}{2\pi} (n_2 - n_1) \]

### Step 4: Analyze Possible Changes
The change in angular momentum can take on values depending on the difference in principal quantum numbers (\( n_2 - n_1 \)). The possible values of \( n_2 - n_1 \) can be integers (0, ±1, ±2, ...), leading to:
- If \( n_2 - n_1 = 1 \), then \( \Delta L = \frac{h}{2\pi} \)
- If \( n_2 - n_1 = 2 \), then \( \Delta L = \frac{2h}{2\pi} = \frac{h}{\pi} \)
- And so on.

### Step 5: Conclusion
Thus, the change in orbital angular momentum during an electron transition inside a hydrogen atom can be:
- \( \Delta L = \frac{h}{2\pi} \) or \( \Delta L = \frac{h}{\pi} \) depending on the transition.

### Final Answer
The orbital angular momentum may change by \( \frac{h}{2\pi} \) or \( \frac{h}{\pi} \).

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