When a hydrogen atoms emits a photon of energy `12.1 eV` , its orbital angular momentum changes by (where h os Planck's constant)

Options

A. `3h/pi`
B. `2h/pi`
C. `h/pi`
D. `4h/pi`

Answer

To solve the problem of how the orbital angular momentum of a hydrogen atom changes when it emits a photon of energy \(12.1 \, \text{eV}\), we can follow these steps:

### Step 1: Understand the Energy Levels of Hydrogen
The energy levels of a hydrogen atom are given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \(n\) is the principal quantum number (1, 2, 3,...).

### Step 2: Set Up the Energy Difference Equation
When a hydrogen atom emits a photon, the energy difference between the two energy levels can be expressed as:
\[ E_2 - E_1 = 12.1 \, \text{eV} \]
Substituting the energy levels:
\[ -\frac{13.6}{n_2^2} - \left(-\frac{13.6}{n_1^2}\right) = 12.1 \]
This simplifies to:
\[ \frac{13.6}{n_1^2} - \frac{13.6}{n_2^2} = 12.1 \]

### Step 3: Factor Out the Common Term
Factoring out \(13.6\) gives:
\[ 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = 12.1 \]
Dividing both sides by \(13.6\):
\[ \frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{12.1}{13.6} \approx 0.888 \]

### Step 4: Use Trial and Error to Find \(n_1\) and \(n_2\)
We need to find integers \(n_1\) and \(n_2\) that satisfy this equation.

1. **Assume \(n_1 = 1\)**:
\[ \frac{1}{1^2} - \frac{1}{n_2^2} = 0.888 \implies 1 - \frac{1}{n_2^2} = 0.888 \implies \frac{1}{n_2^2} = 0.112 \implies n_2^2 \approx 8.93 \implies n_2 \approx 3 \]

2. **Check the values**:
- For \(n_1 = 1\) and \(n_2 = 3\):
\[ \frac{1}{1^2} - \frac{1}{3^2} = 1 - \frac{1}{9} = \frac{8}{9} \approx 0.888 \]
This pair satisfies the equation.

### Step 5: Calculate the Change in Orbital Angular Momentum
The orbital angular momentum \(L\) of an electron in a hydrogen atom is given by:
\[ L = n \frac{h}{2\pi} \]
Thus, the change in angular momentum when transitioning from \(n_1\) to \(n_2\) is:
\[ \Delta L = L_2 - L_1 = \left(n_2 \frac{h}{2\pi}\right) - \left(n_1 \frac{h}{2\pi}\right) = \left(3 \frac{h}{2\pi}\right) - \left(1 \frac{h}{2\pi}\right) = \left(3 - 1\right) \frac{h}{2\pi} = 2 \frac{h}{2\pi} = \frac{h}{\pi} \]

### Final Answer
The change in orbital angular momentum when the hydrogen atom emits a photon of energy \(12.1 \, \text{eV}\) is:
\[ \Delta L = \frac{h}{\pi} \]