The ionisation potential of hydrogen atom is -13.6 eV. An electron in the ground state of a hydrogen atom absorbs a photon of energy 12.75 eV. How many diggerent spectral lines can one expect when the electron make a downward transition

Options

A. 1
B. 4
C. 2
D. 6

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Ionization Potential
The ionization potential of the hydrogen atom is given as -13.6 eV. This value represents the energy of the electron in the ground state (n=1).

### Step 2: Calculate the Final Energy Level After Absorption
The electron in the ground state absorbs a photon of energy 12.75 eV. We need to find the new energy level of the electron after this absorption.

The energy of the electron after absorbing the photon can be calculated as:
\[ E_{final} = E_{initial} + E_{photon} \]
\[ E_{final} = -13.6 \, \text{eV} + 12.75 \, \text{eV} = -0.85 \, \text{eV} \]

### Step 3: Determine the Corresponding Energy Level
The energy of the hydrogen atom at a given energy level \( n \) is given by the formula:
\[ E_n = -\frac{13.6}{n^2} \, \text{eV} \]
We need to find \( n \) such that:
\[ -\frac{13.6}{n^2} = -0.85 \]
Solving for \( n^2 \):
\[ \frac{13.6}{n^2} = 0.85 \]
\[ n^2 = \frac{13.6}{0.85} \approx 16 \]
\[ n = 4 \]
Thus, the electron is excited to the n=4 energy level.

### Step 4: Calculate the Number of Spectral Lines
When the electron makes downward transitions from n=4 to lower energy levels, we can calculate the number of possible spectral lines. The formula for the number of spectral lines from a given level \( n \) is:
\[ \text{Number of lines} = \frac{n(n-1)}{2} \]
Substituting \( n = 4 \):
\[ \text{Number of lines} = \frac{4(4-1)}{2} = \frac{4 \times 3}{2} = 6 \]

### Conclusion
The total number of different spectral lines that can be expected when the electron makes downward transitions is **6**.

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