Question

Taking Rydberg's constant ${R_H} = 1.097 \times {10^7}m$ first and second wavelength of Balmer series in hydrogen spectrum is:
A. $2000\mathop A\limits^o ,\,3000\mathop A\limits^o $
B. $1575\mathop A\limits^o ,\,2960\mathop A\limits^o $
C. $6529\mathop A\limits^o ,\,4280\mathop A\limits^o $
D. $6552\mathop A\limits^o ,\,4863\mathop A\limits^o $

Answer 1

Hint:As we know that the Number of Spectral Lines defines an element's absorption spectrum that has dark lines in the same positions as its emission spectrum's brilliant lines. We will apply Rydberg's formula to compute the second wavelength 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:
$\dfrac{1}{\lambda } = R \times {Z^2}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right){m^{ - 1}}$
Here, $\lambda $ is the wavelength, $R$ is the Rydberg constant and $n_1,n_2$ are the energy states.

Complete step by step solution:
In the question, Rydberg’s constant is given by:
${R_H} = 1.097 \times {10^7}m$
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.

In order to know that the wavelength of a photon released by a hydrogen atom transitioning from its ${n_1}$ to ${n_2}$ state is given by:
$\dfrac{1}{\lambda } = R \times {Z^2}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right){m^{ - 1}}$
Assume that the wavelengths of the first and second lines in the Balmer series are${\lambda _1}$ and ${\lambda _2}$, respectively. For first wavelength of 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:
$\dfrac{1}{{{\lambda _1}}} = 1.097 \times {10^7}\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}} \right)\\$
$\Rightarrow \dfrac{1}{{{\lambda _1}}} = 1.097 \times {10^7}\left( {\dfrac{1}{4} - \dfrac{1}{9}} \right)\\$
$\Rightarrow {\lambda _1} \approx 6563\mathop A\limits^o \\$
Similarly, for the second wavelength of Balmer series, the ground state for the Balmer series is ${n_1} = 2$, the just-next-energy state, which is ${n_2} = 4$, would be the nearest energy state for the least energy,
$\dfrac{1}{{{\lambda _2}}} = 1.097 \times {10^7}\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{4^2}}}} \right)\\$
$\Rightarrow \dfrac{1}{{{\lambda _2}}} = 1.097 \times {10^7}\left( {\dfrac{1}{4} - \dfrac{1}{{16}}} \right) \\$
$\Rightarrow {\lambda _2} \approx 4861\mathop A\limits^o $
The nearest values of ${\lambda _1} \approx 6563\mathop A\limits^o $ and ${\lambda _2} \approx 4861\mathop A\limits^o $ are ${\lambda _1} \approx 6552\mathop A\limits^o $ and ${\lambda _2} \approx 4863\mathop A\limits^o $ respectively.

Hence, the correct option is D.

Additional Information: The highest wavelength of the Balmer series in the hydrogen spectrum corresponds to a transition between the second and third shells, while the minimum wavelength corresponds to a transition between the second and fourth shells.

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.