The electron in a hydrogen atom at rest makes a transition from `n = 2` energy state to the `n = 1` ground state.
find the energy `(eV)` of the emitted photon.

Options

A. `5.8 eV`
B. `8.3 eV`
C. `10.2 eV`
D. `12.7 eV`

Explanation

To find the energy of the emitted photon when an electron in a hydrogen atom transitions from the n = 2 energy state to the n = 1 ground state, we can use the Rydberg formula for hydrogen:

### Step 1: Use the Rydberg formula
The Rydberg formula for the wavelength of the emitted photon during a transition between energy levels is given by:

\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

where:
- \( R \) is the Rydberg constant, approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \)
- \( n_1 \) is the lower energy level (1 for the ground state)
- \( n_2 \) is the higher energy level (2 for the excited state)

### Step 2: Substitute the values into the formula
For our case:
- \( n_1 = 1 \)
- \( n_2 = 2 \)

Substituting these values into the Rydberg formula gives:

\[ \frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]

Calculating the right-hand side:

\[ \frac{1}{\lambda} = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]

\[ \frac{1}{\lambda} = \frac{3R}{4} \]

### Step 3: Calculate the wavelength
Now substituting the value of \( R \):

\[ \frac{1}{\lambda} = \frac{3 \times 1.097 \times 10^7}{4} = \frac{3.291 \times 10^7}{4} = 8.2275 \times 10^6 \, \text{m}^{-1} \]

Now, taking the reciprocal to find \( \lambda \):

\[ \lambda = \frac{1}{8.2275 \times 10^6} \approx 1.215 \times 10^{-7} \, \text{m} = 121.5 \, \text{nm} \]

### Step 4: Calculate the energy of the photon
The energy \( E \) of the photon can be calculated using the formula:

\[ E = \frac{hc}{\lambda} \]

where:
- \( h \) is Planck's constant, approximately \( 6.626 \times 10^{-34} \, \text{Js} \)
- \( c \) is the speed of light, approximately \( 3 \times 10^8 \, \text{m/s} \)

Substituting the values:

\[ E = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{1.215 \times 10^{-7}} \]

Calculating \( E \):

\[ E \approx \frac{1.9878 \times 10^{-25}}{1.215 \times 10^{-7}} \approx 1.635 \times 10^{-18} \, \text{J} \]

### Step 5: Convert energy from Joules to electronvolts
To convert the energy from Joules to electronvolts, we use the conversion factor \( 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \):

\[ E \approx \frac{1.635 \times 10^{-18}}{1.602 \times 10^{-19}} \approx 10.19 \, \text{eV} \]

### Final Answer
The energy of the emitted photon is approximately **10.19 eV**.

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