Question

The first member of the Balmer series of hydrogen atoms has a wavelength of $6563\mathop A\limits^o $. Calculate the wavelength and frequency of the second member of the same series. Given $C = 3 \times {10^8}m{s^{ - 1}}$.

Answer 1

Hint:As we know that when an electron transitions from a higher level down to an energy level, a series of spectrum emission lines called the Balmer series of hydrogen atoms results. We will apply Rydberg's formula to compute the wavelength and frequency of the Balmer series 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.
And, the formula of frequency is represented as:
$\nu = \dfrac{c}{{{\lambda}}}$.
Here, $\nu$ is the frequency of radiation, $c$ is the speed of light and $\lambda $ is the wavelength.

Complete step by step solution:
In the question, we have given the wavelength and velocity of the first member $6563\mathop A\limits^o $ and $C = 3 \times {10^8}m{s^{ - 1}}$ respectively. Now, the transition of the electron must occur in a way that releases the least amount of energy in order for the maximal wavelength to be emitted. Additionally, the transfer of electrons between the two closest energy states should occur for the least energy. As we know that the wavelength of different lines of the Balmer series is given by:
$\dfrac{1}{\lambda } = {R_H}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$
Where ${n_2} = 3,4,5,....$

For the first member of the Balmer series, the ground state for the Balmer series is ${n_1} = 2$, the just-next-energy state, which is ${n_2} = 3$, would be the nearest energy state for the least energy. Now, substituting the values in the above formula, then we obtain:
$\dfrac{1}{{{\lambda _a}}} = {R_H}\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}} \right) \\$
 $\Rightarrow \dfrac{1}{{{\lambda _a}}} = {R_H}\left( {\dfrac{1}{4} - \dfrac{1}{9}} \right) \\$
 $\Rightarrow {\lambda _a} = \dfrac{{36}}{{5{R_H}}}\,\,\,\,\,\,....(i) \\$
Similarly, for the second member of Balmer series, the ground state for the Balmer series is ${n_1} = 1$, the just-next-energy state, which is ${n_2} = 3$, would be the nearest energy state for the least energy,
$\dfrac{1}{{{\lambda _b}}} = {R_H}\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{3^2}}}} \right) \\$
$\Rightarrow \dfrac{1}{{{\lambda _b}}} = {R_H}\left( {1 - \dfrac{1}{9}} \right) \\$
$\Rightarrow {\lambda _b} = \dfrac{9}{{8{R_H}}}\,\,\,\,\,\,....(ii) \\$
Dividing the equation $(i)$ by $(ii)$, then we have:
$\dfrac{{{\lambda _a}}}{{{\lambda _b}}} = \dfrac{{\dfrac{{36}}{{5{R_H}}}}}{{\dfrac{9}{{8{R_H}}}}} \\$
$\Rightarrow \dfrac{{{\lambda _a}}}{{{\lambda _b}}} = \dfrac{{36}}{5} \times \dfrac{8}{9}\\$
$\Rightarrow \dfrac{{{\lambda _a}}}{{{\lambda _b}}} = \dfrac{{32}}{5} \\$
Substitute the value of ${\lambda _a} = 6563$ in the obtained equation, then we obtain:
$\dfrac{{6563 \times {{10}^{ - 10}}}}{{{\lambda _b}}} = \dfrac{{32}}{5} \\$
$\Rightarrow {\lambda _b} = \dfrac{{5 \times 6563}}{{32}} \\$
$\Rightarrow {\lambda _b} = 1025.5\mathop A\limits^o \\$
Now, using the formula of frequency to determine the frequency of the second member $\nu = \dfrac{c}{{{\lambda _b}}}$, then we obtain:
$\nu = \dfrac{{3 \times {{10}^8}}}{{1025.5 \times {{10}^{ - 10}}}} \\$
$\Rightarrow \nu = 0.002925 \times {10^{18}} \\$
$\therefore \nu = 2.925 \times {10^{15}}Hz \\$
Hence, the second member of the Balmer series member's wavelength and frequency are $1025.5\mathop A\limits^o $ and $2.925 \times {10^{15}}Hz$, respectively.

Note: > It should be noted that ${n_2}$ 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 and a single photon's energy is directly proportional to its electromagnetic frequency, and as a result, it is inversely proportional to its wavelength.