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

Answer

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

### Step 1: Calculate the energy of the electron in the n = 3 state.
The formula for the energy of a hydrogen atom is given by:

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

Substituting \( n = 3 \):

\[ E_3 = \frac{-13.6}{3^2} = \frac{-13.6}{9} \approx -1.51 \text{ eV} \]

### Step 2: Calculate the energy of the electron in the n = 2 state.
Now, we substitute \( n = 2 \):

\[ E_2 = \frac{-13.6}{2^2} = \frac{-13.6}{4} = -3.4 \text{ eV} \]

### Step 3: Calculate the energy difference (ΔE) between the two states.
The energy difference when the electron jumps from n = 3 to n = 2 is given by:

\[ \Delta E = E_2 - E_3 \]

Substituting the values we calculated:

\[ \Delta E = (-3.4) - (-1.51) = -3.4 + 1.51 = -1.89 \text{ eV} \]

Since we are interested in the energy of the photon ejected, we take the absolute value:

\[ \Delta E \approx 1.89 \text{ eV} \]

### 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 **1.89 eV**.

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