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 follow these steps:

### Step 1: Use the Rydberg Formula
The Rydberg formula for the wavelength of emitted or absorbed light during electron transitions in hydrogen is given by:

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

Where:
- \( R \) is the Rydberg constant, approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \)
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \))
- \( n_1 \) is the lower energy level (1 for ground state)
- \( n_2 \) is the higher energy level (2 in this case)

### Step 2: Substitute the Values
Substituting the values into the Rydberg formula:

\[ \frac{1}{\lambda} = 1.097 \times 10^7 \cdot 1^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]

Calculating the right side:

\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( 1 - \frac{1}{4} \right) = 1.097 \times 10^7 \left( \frac{3}{4} \right) \]

\[ \frac{1}{\lambda} = 1.097 \times 10^7 \cdot 0.75 = 0.823 \times 10^7 \, \text{m}^{-1} \]

### Step 3: Calculate the Wavelength
Now, take the reciprocal to find \( \lambda \):

\[ \lambda = \frac{1}{0.823 \times 10^7} \approx 1.215 \times 10^{-7} \, \text{m} \]

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

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

Where:
- \( h \) is Planck's constant, approximately \( 6.626 \times 10^{-34} \, \text{J s} \)
- \( c \) is the speed of light, approximately \( 3.00 \times 10^8 \, \text{m/s} \)

Substituting the values:

\[ E = \frac{(6.626 \times 10^{-34}) \cdot (3.00 \times 10^8)}{1.215 \times 10^{-7}} \]

Calculating this gives:

\[ E \approx \frac{1.9878 \times 10^{-25}}{1.215 \times 10^{-7}} \approx 1.63 \times 10^{-18} \, \text{J} \]

### Step 5: Convert Energy to Electron Volts
To convert joules to electron volts, use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):

\[ E \approx \frac{1.63 \times 10^{-18}}{1.6 \times 10^{-19}} \approx 10.19 \, \text{eV} \]

### Final Answer
The energy of the emitted photon is approximately **10.2 eV**.

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