Step-by-step Solution:

1. Calculate the energy of the incident photon. The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$. Given values are:

   • Wavelength of light, $\lambda = 970 \mathop A\limits^o = 970 \times 10^{-10} \text{ m}$.
   • Product of Planck's constant and speed of light, $hc = 1.237 \times 10^{-6} \text{ eVm}$.

   Substitute these values into the energy formula: $E_{photon} = \frac{1.237 \times 10^{-6} \text{ eVm}}{970 \times 10^{-10} \text{ m}} = \frac{1.237 \times 10^{-6}}{9.7 \times 10^{-8}} \text{ eV}$ $E_{photon} = \frac{1.237}{9.7} \times 10^2 \text{ eV} = \frac{123.7}{9.7} \text{ eV} \approx 12.75 \text{ eV}$

2. Determine the excited state of the hydrogen atom. The hydrogen atom is initially in its ground state ($n=1$), which has an energy of $E_1 = -13.6 \text{ eV}$. When the atom absorbs the photon, the electron transitions to a higher energy level, $n$. The energy of the electron in this new state, $E_n$, is the sum of its initial energy and the absorbed photon's energy. $E_n = E_1 + E_{photon}$ $E_n = -13.6 \text{ eV} + 12.75 \text{ eV} = -0.85 \text{ eV}$

   The energy of an electron in the n-th state of a hydrogen atom is given by the formula $E_n = \frac{-13.6}{n^2} \text{ eV}$. We can use this to find the principal quantum number 'n' of the excited state. $-0.85 \text{ eV} = \frac{-13.6}{n^2} \text{ eV}$ $n^2 = \frac{-13.6}{-0.85} = \frac{1360}{85}$ To simplify the fraction, divide the numerator and denominator by 5: $n^2 = \frac{272}{17} = 16$ Taking the square root, we get: $n = \sqrt{16} = 4$ So, the electron is excited to the $n=4$ energy level.

3. Calculate the number of lines in the emission spectrum. After excitation to the $n=4$ state, the electron will de-excite to lower energy levels, emitting photons in the process. Each possible transition corresponds to a spectral line. The total number of possible spectral lines when an electron de-excites from the n-th level is given by the formula: $N = \frac{n(n-1)}{2}$ For $n=4$, the number of lines is: $N = \frac{4(4-1)}{2} = \frac{4 \times 3}{2} = \frac{12}{2} = 6$

   The possible transitions are from $n=4$ to $n=3,2,1$; from $n=3$ to $n=2,1$; and from $n=2$ to $n=1$. This gives a total of $3 + 2 + 1 = 6$ lines.

Final Answer:

The number of lines present in the emission spectrum is 6.