Energy `E` of a hydrogen atom with principle quantum number `n` is given by `E = (-13.6)/(n^(2)) eV`. The energy of a photon ejected when the electron jumps from `n = 3` state to `n = 2` state of hydrogen is approximately

Options

A. `1.5eV`
B. `0.85eV`
C. `3.4eV`
D. `1.9eV`

Explanation

To find the energy of the photon ejected when an electron transitions from the n = 3 state to the n = 2 state in a hydrogen atom, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Energy Formula**: The energy \( E \) of a hydrogen atom with principal quantum number \( n \) is given by the formula:
\[ E = \frac{-13.6}{n^2} \text{ eV} \]

2. **Calculate Energy for n = 2**: Substitute \( n = 2 \) into the formula to find the energy of the hydrogen atom in this state:
\[ E_2 = \frac{-13.6}{2^2} = \frac{-13.6}{4} = -3.4 \text{ eV} \]

3. **Calculate Energy for n = 3**: Now, substitute \( n = 3 \) into the formula to find the energy of the hydrogen atom in this state:
\[ E_3 = \frac{-13.6}{3^2} = \frac{-13.6}{9} \approx -1.51 \text{ eV} \]

4. **Find the Energy Difference**: The energy of the photon emitted when the electron transitions from \( n = 3 \) to \( n = 2 \) is the difference in energy between these two states:
\[ E_{\text{photon}} = E_2 - E_3 \]
Substituting the values we calculated:
\[ E_{\text{photon}} = (-3.4) - (-1.51) = -3.4 + 1.51 = -1.89 \text{ eV} \]

5. **Consider the Magnitude**: The energy of the photon is typically expressed as a positive value, so we take the magnitude:
\[ |E_{\text{photon}}| \approx 1.89 \text{ eV} \]

6. **Approximate the Result**: The calculated energy is approximately \( 1.9 \text{ eV} \).

### Final Answer:
The energy of the photon ejected when the electron jumps from the \( n = 3 \) state to the \( n = 2 \) state of hydrogen is approximately \( 1.9 \text{ eV} \).