Energy of the lowest level of hydrogen atom is `-13.6 eV`. The energy of the photon emitted in the transition from n=3 to n=1 is

Options

A. 27 eV
B. 9 eV
C. 3 eV
D. 12.09 eV

Answer

To find the energy of the photon emitted during the transition from n=3 to n=1 in a hydrogen atom, we can follow these steps:

### Step 1: Understand the Energy Levels
The energy of an electron in the nth level of a hydrogen atom is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number.

### Step 2: Calculate the Energy at n=3
Using the formula, we can calculate the energy of the electron at \( n=3 \):
\[ E_3 = -\frac{13.6 \, \text{eV}}{3^2} \]
\[ E_3 = -\frac{13.6 \, \text{eV}}{9} \]
\[ E_3 = -1.51 \, \text{eV} \]

### Step 3: Calculate the Energy at n=1
Next, we calculate the energy of the electron at \( n=1 \):
\[ E_1 = -\frac{13.6 \, \text{eV}}{1^2} \]
\[ E_1 = -13.6 \, \text{eV} \]

### Step 4: Find the Energy of the Photon Emitted
The energy of the photon emitted during the transition from \( n=3 \) to \( n=1 \) is the difference in energy between these two levels:
\[ E_{\text{photon}} = E_1 - E_3 \]
Substituting the values we calculated:
\[ E_{\text{photon}} = (-13.6 \, \text{eV}) - (-1.51 \, \text{eV}) \]
\[ E_{\text{photon}} = -13.6 \, \text{eV} + 1.51 \, \text{eV} \]
\[ E_{\text{photon}} = -12.09 \, \text{eV} \]

Since energy is typically expressed as a positive value for emitted photons, we take the absolute value:
\[ E_{\text{photon}} = 12.09 \, \text{eV} \]

### Final Answer
The energy of the photon emitted in the transition from \( n=3 \) to \( n=1 \) is \( 12.09 \, \text{eV} \).

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