Question

A gas of hydrogen-like ions is prepared in a particular excited state $A$. It emits photons having a wavelength equal to the wavelength of the first line of the Lyman series together with photons of five other wavelengths. Identify the gas and find the principal quantum number of the state$A$.

Answer 1

Hint: The ultraviolet spectrum contains the Lyman series. The principles of the hydrogen spectrum and the number of spectral lines in a given series represent all potential transitions from an energy level that must be applied in order to approach this problem. Atomic hydrogen frequently exhibits an emission spectrum. This spectrum has a propensity to encompass multiple spectral series.

Formula used:
Number of wavelengths emitted in ${n^{th}}$ excited state of an atom can be given as:
$\text{Number of photons emitted}= \dfrac{{n(n - 1)}}{2}$.
Here, $n$ as the principal quantum number.

Complete step by step solution:
In order to know that the quantum of electromagnetic radiation is called a photon. The lowest fundamental unit of a number or the smallest distinct sum of anything is referred to as a quantum. As a result, a photon is one quantum of electromagnetic radiation. Quanta is the plural form of quantum.

The concepts of photons and quanta are derived from quantum theory and quantum mechanics. It suggests that matter and energy are quantized, or that they exist in thin, distinct packets of the smallest possible size. A photon spreads at the speed of light, which means, the total $6$ photons are emitted.

Use the formula of the number of wavelengths that are emitted in ${n^{th}}$ excited state of an atom,
$\text{Number of photons emitted} = \dfrac{{n(n - 1)}}{2}$
Now, substitute the number of photons emitted in the above formula, then we have:
$6 = \dfrac{{n(n - 1)}}{2} \\$
$\Rightarrow 12 = {n^2} - n \\$
$\Rightarrow {n^2} - 4n + 3n - 12 = 0 \\$
$\therefore n = 4,\,n = - 3 $
The value of $n$ is positive instead of negative, so we obtain the number of wavelengths:
$n = 4$.

Therefore, the principal quantum number is $4$and the gas in the ${4^{th}}$excited state.

Note: It should be noted that the wavelength and photons are identical since the energy is directly proportional to the photon's electric frequency. As the wavelength increases, the photon's energy rises as well. With an increase in photon wavelength, the photon's energy diminishes.