A hydrogen atom in the ground state. To which level will it get excited on absorbing a photon of energy 12.75eV `(1eV=1.6 xx 10^(-19))`

Answer

To determine the energy level to which a hydrogen atom in the ground state will get excited upon absorbing a photon of energy 12.75 eV, we can follow these steps:

### Step 1: Convert the energy from eV to joules
The energy of the photon is given in electron volts (eV). We need to convert this energy into joules using the conversion factor \(1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}\).

\[ E = 12.75 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 2.04 \times 10^{-18} \text{ J} \]

### Step 2: Identify the initial energy level
The hydrogen atom is in the ground state, which corresponds to the principal quantum number \(n_1 = 1\).

### Step 3: Use the energy transition formula
The energy difference between two levels in a hydrogen atom is given by the formula:

\[ \Delta E = -\frac{13.6 \, Z^2}{n_{\text{upper}}^2} + \frac{13.6 \, Z^2}{n_{\text{lower}}^2} \]

For hydrogen, \(Z = 1\), and since the atom is initially in the ground state, we have \(n_{\text{lower}} = 1\).

Thus, we can rewrite the equation as:

\[ \Delta E = -\frac{13.6}{n_{\text{upper}}^2} + 13.6 \]

### Step 4: Set the energy absorbed equal to the energy transition
We know the energy absorbed is 12.75 eV, so we set:

\[ 12.75 = 13.6 - \frac{13.6}{n_{\text{upper}}^2} \]

### Step 5: Rearranging the equation
Rearranging the equation gives:

\[ \frac{13.6}{n_{\text{upper}}^2} = 13.6 - 12.75 \]

Calculating the right side:

\[ \frac{13.6}{n_{\text{upper}}^2} = 0.85 \]

### Step 6: Solve for \(n_{\text{upper}}^2\)
Now, we can solve for \(n_{\text{upper}}^2\):

\[ n_{\text{upper}}^2 = \frac{13.6}{0.85} \]

Calculating this gives:

\[ n_{\text{upper}}^2 \approx 16 \]

### Step 7: Find \(n_{\text{upper}}\)
Taking the square root gives:

\[ n_{\text{upper}} = 4 \]

### Final Answer
Thus, the hydrogen atom will get excited to the energy level \(n = 4\).

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