Animated Solution for Physics - Atoms and Nuclei: A hydrogen atom in its ground state is irradiated by light of wavelength $970\text{ A˚}$. Taking $hc/e=1.237\times 10^{-6}\text{ eVm}$ and the ground state energy of hydrogen atom as $-13.6\text{ eV}$, the number of lines present in the emission spectrum is

Imagine a hydrogen atom resting peacefully in its ground state. The electron is in the innermost orbit, tightly bound to the nucleus. According to Bohr's model, the energy of this ground state is exactly $-13.6\text{ eV}$. The negative sign is crucial here—it tells us that the electron is trapped in the electrostatic potential well of the proton. It owes the universe $13.6\text{ eV}$ of energy before it can break free.

To wake this electron up, we irradiate the atom with a specific wavelength of light: $\lambda =970\text{ A˚}$. But how much energy does this incoming photon carry? We use the fundamental Planck-Einstein relation:

The problem generously provides the value of $hc/e=1.237\times 10^{-6}\text{ eVm}$. To keep our units perfectly consistent, we must convert the wavelength from Angstroms to meters by multiplying by $10^{-10}$.

The electron absorbs this entire $12.75\text{ eV}$ energy packet in a single, indivisible quantum event. It jumps to a higher energy level. Its new total energy becomes the sum of its initial bound energy and the absorbed photon energy:

Now, which specific orbit corresponds to this new energy? We return to Bohr's energy formula for the $n$-th orbit:

Equating the two expressions for $E_n$:

The electron has reached the third excited state ($n=4$)!

Finally, what goes up must come down. As the electron falls back to the ground state, it doesn't have to take a single leap. It can cascade down through intermediate energy levels, emitting a photon at each step. The total number of possible spectral lines is a classic combinations problem—we are choosing any 2 levels out of the $n$ available levels. The formula is:

Plugging in $n=4$:

Out of these 6 lines, 3 end at the ground state (Lyman series), 2 end at $n=2$ (Balmer series), and 1 ends at $n=3$ (Paschen series). The physics is as elegant as it is precise!