Ionisation potential of hydrogen atom is 13.6 eV. Hydrogen atoms on the ground state are rarely excited by monochromatic radiation of photon 12.1 eV. The special line emitted by a hydrogen atom according to Bohr’s theory will be:
A. One
B. Two
C. Three
D. Four

Hint: First we will calculate to what level will an electron excite by the given radiation. Even though it is given in the question that it rarely gets excited by this radiation, we will assume that hydrogen atom gets excited to this level and we will calculate how many lines will be emitted by hydrogen atom.

Complete answer:
If we take the energy level of an electron in its ground state, the total energy of the electron is given as 13.6eV. This energy is negative i.e. it is the sum of attractive potential energy and the kinetic energy of the electron. When excited by a radiation of 12.1 eV, its energy would become $-13.6+12.1=-1.5$eV. When we take the ratio of these two energies of electrons, we get $\dfrac{13.6}{1.5}=9$. This ratio is the square of the orbit number in which the electron will now be. So, an electron reaches the ${{3}^{rd}}$ orbital by radiation of this energy. Number of spectral lines emitted according to Bohr’s theory will be given by the formula $\dfrac{n(n-1)}{2}$, if we use the value of n as 3 we get that there will be a total of $\dfrac{3(3-1)}{2}=3$ spectral lines. Hence, the correct option is C, i.e. Three.

Note:
Take care that the first excitation level is actually the second orbit for a Hydrogen electron. The first orbit is referred to as the ground state. Also, the energy of an electron in orbit is negative that is why they excite when provided with energy and their energy reduces.