Question

Whenever a photon is emitted by hydrogen in the Balmer series, it is followed by another photon in the Lyman series. What wavelength does this latter photon correspond to?

Answer 1

Hint:As we know that the hydrogen atom's electrons constantly transition from one higher energy level to another in order to de-excite themselves. These transitions are separated into five separate series. An electron emits a photon of wavelength-specific radiation during these transitions. We will apply Rydberg's formula to compute the wavelength in this solution. In the Balmer series, the maximum wavelength is obtained when the minimal energy transition occurs, that is, when the electron transition occurs on the second line.

Formula used:
Rydberg’s Formula is as follows:
$\dfrac{1}{\lambda } = {R_H}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$
Here, $\lambda $ is the wavelength, $R_H$ is the Rydberg constant and $n_1,n_2$ are the energy states.

Complete step by step solution:
In order to know that the hydrogen radiation spectrum is divided into several series. Each series is given its discoverer's name. These series include the Balmer series, Pfund series, Paschen series, Brackett, and Lyman. The wavelength of radiations released from hydrogen atoms, according to Bohr's hypothesis, is given by;
$\dfrac{1}{\lambda } = {R_H}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$

In the Lyman series, ${n_2} = 2,3,4,.....$, and ${n_1} = 1$. The photon moves from an energy level higher than its own, such as the second, third, or fourth level, to a level lower than its own, which is the first atomic level. While the Balmer series or radiation is represented by the numbers ${n_1} = 2$ and ${n_2} = 3,4,5.....$.

From the given information, the Lyman series followed by the Balmer series, that means, from $n = 2$ to $n = 1$. As we know, a physical constant related to the electromagnetic spectrum of an atom is called the Rydberg's constant. The Rydberg's constant, which was required for the Rydberg formula for the hydrogen spectral series, came from an empirical fitting parameter. The Rydberg's constant is equal to $1.09678 \times {10^7}{m^{ - 1}} \approx 1.097 \times {10^7}{m^{ - 1}}$.

Now, substitute the value of ${n_1} = 2,\,{n_2} = 1$ and ${R_H} = 1.097 \times {10^7}$ in the above formula, then we obtain:
$\dfrac{1}{\lambda } = 1.097 \times {10^7}\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{1^2}}}} \right) \\$
$\Rightarrow \dfrac{1}{\lambda } = 1.097 \times {10^7}\left( {\dfrac{1}{4} - \dfrac{1}{1}} \right) \\$
$\Rightarrow \lambda = \dfrac{4}{{1.097 \times 3}} \times {10^7} \\$
$\therefore \lambda \approx 122\,nm $

Hence, later photons' wavelengths will be $122\,nm$.

Note: It should be noted that on the basis of Bohr's atomic model, the wavelength was calculated above using Rydberg's formula and Rydberg's constant. And,${n_2}$ is a higher shell and ${n_1}$ is a lower shell when using Rydberg's Formula. If both wave numbers or wavelengths are switched, the result will be negative, which is not conceivable. The atomic spectrum of hydrogen and other atoms as well as ions are explained by this paradigm.