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 find the change in orbital angular momentum (\( \Delta L \)) of a hydrogen atom that emits a photon of energy \( 12.1 \, \text{eV} \), we can follow these steps:

### Step 1: Identify the energy levels involved
The energy of the emitted photon corresponds to a transition between two energy levels in the hydrogen atom. The energy difference between these levels is given by:

\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]

For a photon energy of \( 12.1 \, \text{eV} \), we can find the initial and final energy levels. The transition occurs from \( n = 3 \) to \( n = 1 \):

\[ E_3 - E_1 = 12.1 \, \text{eV} \]

### Step 2: Calculate the change in orbital angular momentum
The orbital angular momentum (\( L \)) of an electron in a hydrogen atom is given by:

\[ L = n \frac{h}{2\pi} \]

where \( n \) is the principal quantum number and \( h \) is Planck's constant.

Now, we can calculate the change in angular momentum (\( \Delta L \)) as follows:

\[ \Delta L = L_1 - L_3 = \left(1 \cdot \frac{h}{2\pi}\right) - \left(3 \cdot \frac{h}{2\pi}\right) \]

This simplifies to:

\[ \Delta L = \left(1 - 3\right) \frac{h}{2\pi} = -2 \frac{h}{2\pi} = -\frac{h}{\pi} \]

### Step 3: Substitute the value of Planck's constant
Using \( h \approx 6.626 \times 10^{-34} \, \text{J s} \):

\[ \Delta L = -\frac{6.626 \times 10^{-34}}{\pi} \approx -2.11 \times 10^{-34} \, \text{J s} \]

### Final Result
Thus, the change in orbital angular momentum \( \Delta L \) is:

\[ \Delta L \approx 2.11 \times 10^{-34} \, \text{J s} \]

(Note: The negative sign indicates a decrease in angular momentum.)

---