The ionisation potential of `H`-atom is `13.6 eV`. When it is excited from ground state by monochromatic radiations of `970.6 Å`, the number of emission lines will be (according to Bohr's theory)

Options

A. `10`
B. `8`
C. `6`
D. `4`

Explanation

To solve the problem, we need to determine the number of emission lines produced when a hydrogen atom is excited from its ground state by monochromatic radiation of wavelength 970.6 Å. We will use Bohr's theory to find the energy levels and the number of possible transitions.

### Step 1: Convert the wavelength to meters
The wavelength given is 970.6 Å. We need to convert this to meters for calculations:

\[ 970.6 \, \text{Å} = 970.6 \times 10^{-10} \, \text{m} \]

### Step 2: Calculate the energy of the incoming photon
Using the formula for energy of a photon:

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

Where:
- \( h = 6.626 \times 10^{-34} \, \text{Js} \) (Planck's constant)
- \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light)
- \( \lambda = 970.6 \times 10^{-10} \, \text{m} \)

Substituting the values:

\[ E = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{970.6 \times 10^{-10}} \]

Calculating this gives:

\[ E \approx 2.06 \times 10^{-19} \, \text{J} \]

To convert this energy into electron volts (eV), we use the conversion \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):

\[ E \approx \frac{2.06 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 1.2875 \, \text{eV} \]

### Step 3: Determine the final energy level (n2)
The ionization potential of the hydrogen atom is given as 13.6 eV, which corresponds to the energy of the ground state (n=1). The energy levels of the hydrogen atom are given by:

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

The energy required to excite the atom from n=1 to n=n2 is:

\[ E_{excitation} = E_{n2} - E_{1} = -\frac{13.6}{n2^2} - (-13.6) \]

Setting this equal to the energy of the photon:

\[ -\frac{13.6}{n2^2} + 13.6 = 1.2875 \]

Rearranging gives:

\[ -\frac{13.6}{n2^2} = 1.2875 - 13.6 \]

Calculating the right side:

\[ -\frac{13.6}{n2^2} = -12.3125 \]

Thus:

\[ \frac{13.6}{n2^2} = 12.3125 \]

Solving for \( n2^2 \):

\[ n2^2 = \frac{13.6}{12.3125} \approx 1.103 \]

Taking the square root gives:

\[ n2 \approx 4 \]

### Step 4: Calculate the number of emission lines
The number of emission lines is given by the formula:

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

Where \( n \) is the final energy level (n2). Here, \( n2 = 4 \):

\[ \text{Number of lines} = \frac{4(4-1)}{2} = \frac{4 \times 3}{2} = 6 \]

### Final Answer
The number of emission lines will be **6**.

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