The hydrogen atom in ground state is excited by a monochromatic radiation of wavelength `lambda = 1025 A`. The possible number of emmision spectral lines in the spectrum will be (hc = 12400 eV A)

Options

A. 4
B. 6
C. 5
D. 3

Answer

To solve the problem of determining the possible number of emission spectral lines in the spectrum of a hydrogen atom excited by monochromatic radiation of wavelength \( \lambda = 1025 \) Å, we will follow these steps:

### Step 1: Calculate the energy of the incoming radiation
The energy \( E \) of the photon can be calculated using the formula:

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

Given:
- \( hc = 12400 \, \text{eV} \cdot \text{Å} \)
- \( \lambda = 1025 \, \text{Å} \)

Substituting the values:

\[ E = \frac{12400 \, \text{eV} \cdot \text{Å}}{1025 \, \text{Å}} \approx 12.09 \, \text{eV} \]

### Step 2: Determine the final energy level \( n_2 \)
The energy levels of the hydrogen atom are given by:

\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]

For the ground state (\( n_1 = 1 \)), the energy is:

\[ E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV} \]

The change in energy when the atom is excited to a higher level \( n_2 \) is:

\[ \Delta E = E_{n_2} - E_1 = 12.09 \, \text{eV} \]

Substituting for \( E_{n_2} \):

\[ E_{n_2} - (-13.6) = 12.09 \]

This simplifies to:

\[ E_{n_2} = 12.09 - 13.6 = -1.51 \, \text{eV} \]

Setting this equal to the energy level equation:

\[ -\frac{13.6}{n_2^2} = -1.51 \]

Rearranging gives:

\[ \frac{13.6}{1.51} = n_2^2 \]

Calculating:

\[ n_2^2 \approx 9.01 \implies n_2 \approx 3 \]

### Step 3: Determine the possible transitions
The possible transitions from the excited state \( n_2 \) back to the ground state \( n_1 \) can be calculated using the formula for the number of spectral lines:

\[ \text{Number of lines} = \frac{n(n-1)}{2} \]

Where \( n \) is the number of energy levels involved. Since we can have transitions from \( n = 2 \) to \( n = 3 \), we consider \( n = 3 \):

\[ \text{Number of lines} = \frac{3(3-1)}{2} = \frac{3 \times 2}{2} = 3 \]

### Conclusion
Thus, the possible number of emission spectral lines in the spectrum will be **3**.

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