If a hydrogen atom emit a photon of energy `12.1 eV` , its orbital angular momentum changes by `Delta L. then`Delta L` equals

Options

A. `1.05 xx 10^(-34) J s`
B. `2.11 xx 10^(-34) J s`
C. `3.16 xx 10^(-34) J s`
D. `4.22 xx 10^(-34)`

Explanation

To solve the problem of finding the change in orbital angular momentum (ΔL) when a hydrogen atom emits a photon of energy 12.1 eV, we can follow these steps:

### Step 1: Understand the Energy Levels of Hydrogen Atom
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.

### Step 2: Calculate the Initial and Final Energy Levels
We need to find the initial and final energy levels corresponding to the emission of a photon with energy 12.1 eV.

1. The energy of the photon emitted is:
\[ E_{\text{photon}} = 12.1 \, \text{eV} \]

2. The transition occurs from a higher energy level \( E_i \) to a lower energy level \( E_f \). Therefore:
\[ E_i - E_f = E_{\text{photon}} \]

3. We will calculate the energy levels for \( n = 1, 2, 3, 4 \):
- For \( n = 1 \):
\[ E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV} \]
- For \( n = 2 \):
\[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -3.4 \, \text{eV} \]
- For \( n = 3 \):
\[ E_3 = -\frac{13.6 \, \text{eV}}{3^2} = -1.51 \, \text{eV} \]
- For \( n = 4 \):
\[ E_4 = -\frac{13.6 \, \text{eV}}{4^2} = -0.85 \, \text{eV} \]

### Step 3: Determine the Transition
From the calculated energy levels, we can see that:
- The transition must be from \( n = 3 \) to \( n = 1 \) because:
\[ E_3 - E_1 = (-1.51) - (-13.6) = 12.1 \, \text{eV} \]

### Step 4: Calculate the Change in Orbital Angular Momentum (ΔL)
The change in orbital angular momentum is given by:
\[ \Delta L = L_i - L_f = (n_i - n_f) \cdot \frac{h}{2\pi} \]
where \( n_i = 3 \) and \( n_f = 1 \).

1. Substitute the values:
\[ \Delta L = (3 - 1) \cdot \frac{h}{2\pi} = 2 \cdot \frac{h}{2\pi} = \frac{h}{\pi} \]

2. Using Planck's constant \( h = 6.626 \times 10^{-34} \, \text{J s} \):
\[ \Delta L = \frac{6.626 \times 10^{-34}}{\pi} \approx 2.11 \times 10^{-34} \, \text{J s} \]

### Final Answer
Thus, the change in orbital angular momentum \( \Delta L \) is approximately:
\[ \Delta L \approx 2.11 \times 10^{-34} \, \text{J s} \]

---