The ionisation energy of hydrogen is 13.6 eV . The energy of the photon released when an electron jumps from the first excited state (n=2) to the ground state of hydrogen atom is

Options

A. 3.4 eV
B. 4.53 eV
C. 10.2 eV
D. 13.6eV

Explanation

To find the energy of the photon released when an electron jumps from the first excited state (n=2) to the ground state (n=1) of a hydrogen atom, we can follow these steps:

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

### Step 2: Calculate the Energy at n=2
For the first excited state (n=2):
\[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \]

### Step 3: Calculate the Energy at n=1
For the ground state (n=1):
\[ E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV} \]

### Step 4: Calculate the Energy of the Photon Released
The energy of the photon released during the transition from n=2 to n=1 is the difference in energy between these two states:
\[ E_{\text{photon}} = E_1 - E_2 \]
Substituting the values we calculated:
\[ E_{\text{photon}} = (-13.6 \, \text{eV}) - (-3.4 \, \text{eV}) = -13.6 \, \text{eV} + 3.4 \, \text{eV} = -10.2 \, \text{eV} \]
Since we are interested in the magnitude of the energy released, we take the positive value:
\[ E_{\text{photon}} = 10.2 \, \text{eV} \]

### Final Answer
The energy of the photon released when an electron jumps from the first excited state (n=2) to the ground state (n=1) of a hydrogen atom is **10.2 eV**.

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