A hydrogen atom in its ground state is irradiated by light of wavelength `970Å` Taking `hc//e = 1.237 xx 10^(-6)`eV m and the ground state energy of hydrogen atom as ` - 13.6 eV` the number of lines present in the emmission spectrum is

Explanation

To solve the problem step by step, we will follow the concepts of energy levels in a hydrogen atom and the relationship between the energy of a photon and its wavelength.

### Step 1: Calculate the Energy of the Photon
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h \) is Planck's constant multiplied by the speed of light, given as \( 1.237 \times 10^{-6} \, \text{eV m} \)
- \( \lambda \) is the wavelength of the light in meters. Since \( 970 \, \text{Å} = 970 \times 10^{-10} \, \text{m} \), we convert it to meters.

Now substituting the values:
\[ E = \frac{1.237 \times 10^{-6}}{970 \times 10^{-10}} \text{ eV} \]
Calculating this gives:
\[ E \approx 12.7 \, \text{eV} \]

### Step 2: Determine the Final Energy Level
The ground state energy of the hydrogen atom is given as \( -13.6 \, \text{eV} \). When the photon is absorbed, the total energy of the atom becomes:
\[ E_{\text{final}} = E_{\text{ground}} + E_{\text{photon}} = -13.6 \, \text{eV} + 12.7 \, \text{eV} \]
Calculating this gives:
\[ E_{\text{final}} = -0.9 \, \text{eV} \]

### Step 3: Relate the Final Energy to the Energy Level Formula
The energy of the hydrogen atom at level \( n \) is given by:
\[ E_n = -\frac{Z^2 \cdot 13.6}{n^2} \text{ eV} \]
For hydrogen, \( Z = 1 \), so:
\[ E_n = -\frac{13.6}{n^2} \]
Setting \( E_n = -0.9 \, \text{eV} \):
\[ -0.9 = -\frac{13.6}{n^2} \]
This simplifies to:
\[ n^2 = \frac{13.6}{0.9} \approx 15.11 \]
Taking the square root gives:
\[ n \approx 4 \]

### Step 4: Calculate the Number of Emission Lines
When the electron transitions from the \( n = 4 \) level to lower levels, the number of possible emission lines can be calculated using the formula:
\[ \text{Number of lines} = \frac{n(n-1)}{2} \]
Substituting \( n = 4 \):
\[ \text{Number of lines} = \frac{4 \times 3}{2} = 6 \]

### Final Answer
Thus, the number of lines present in the emission spectrum is:
\[ \boxed{6} \]

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