Question

The wavelength of $ {H_\alpha } $ line Balmer’s series of hydrogen spectrum is $ 6563\mathop A\limits^ \circ $ . Find the:
(A) Wavelength of $ {H_\gamma } $ (gamma) line of the Balmer series.
(B) Shortest wavelength of Brackett series

Answer 1

Hint The alpha line of the Balmer series corresponds to a de-excitation from the third energy level to the second energy level while the gamma line is from the fifth energy level to the second energy level.
The shortest wavelength of any series corresponds to the line which emits the highest energy.

Formula used: $ \dfrac{1}{\lambda } = {R_H}\left( {\dfrac{1}{{n_f^2}} - \dfrac{1}{{n_i^2}}} \right) $ where $ \lambda $ is wavelength, $ {n_i} $ is the initial energy level of the electron, $ {n_f} $ is the final energy level of the electron and is the Rydberg constant.

Complete step by step answer
There are many series of hydrogen atoms, and each series has a base energy level. Each line of the series is the de-excitation of an electron from a higher energy level to the base energy level for that series. For the Balmer’s series the base energy level is the second principal quantum number, 2. For Bracket series, it is 4.
The $ {H_\alpha } $ line Balmer’s series corresponds to a de-excitation from $ n = 3 $ to $ n = 2 $ .
Hence,
$\Rightarrow \dfrac{1}{{{\lambda _\alpha }}} = {R_H}\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}} \right) $ where $ {\lambda _\alpha } $ is the wavelength of $ {H_\alpha } $
According to the equation, this is equal to $ 6563\mathop A\limits^ \circ = 6.563 \times {10^{ - 7}}m $ since $ 1\mathop A\limits^ \circ = {10^{ - 10}}m $ .
Thus, inserting the value into equation above we have
$\Rightarrow \dfrac{1}{{6.563 \times {{10}^{ - 7}}}} = {R_H}\left( {\dfrac{1}{4} - \dfrac{1}{9}} \right) = {R_H}\left( {\dfrac{{9 - 4}}{{36}}} \right) $
By calculating for $ {R_H} $ , we have
$\Rightarrow {R_H} = 1.0972 \times {10^7}{m^{ - 1}} $
Also, for $ {H_\gamma } $ for which de-excitation occurs from $ n = 5 $ to $ n = 2 $ we have
$\Rightarrow \dfrac{1}{{{\lambda _\gamma }}} = {R_H}\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{5^2}}}} \right) $ where $ {\lambda _\gamma } $ is the wavelength of $ {H_\gamma } $
With our value for $ {R_H} $ and simplifying the bracket, we have
$\Rightarrow \dfrac{1}{{{\lambda _\gamma }}} = 1.0972 \times {10^7}\left( {\dfrac{1}{4} - \dfrac{1}{{25}}} \right) = 1.0972 \times {10^7}\left( {\dfrac{{21}}{{100}}} \right) $
$\Rightarrow \dfrac{1}{{{\lambda _\gamma }}} = 2.304120 \times {10^6} $
Hence, by inverting both sides,
 $\Rightarrow {\lambda _\gamma } = 4.340 \times {10^{ - 7}}m $
The shortest wavelength of a series corresponds to the de-excitation that emits the highest energy (because $ E \propto \dfrac{1}{\lambda } $ ). The energy of an electron increases as it goes from a lower energy level to a higher energy level by absorption of energy. Hence, as the electron de-excites, it releases the energy equal to the energy that was previously absorbed. An electron absorbs the most energy when it gets excited to infinity, thus also releases the most as it de-excites from infinity. For the Bracket series with a base level of $ n = 4 $ , the highest release of energy is from $ n = \infty $ to $ n = 4 $ .
Hence,
$\Rightarrow \dfrac{1}{{{\lambda _s}}} = {R_H}\left( {\dfrac{1}{{{4^2}}} - \dfrac{1}{\infty }} \right) $ where $ {\lambda _s} $ is the shortest wavelength of the Bracket series.
$\Rightarrow \dfrac{1}{{{\lambda _s}}} = {R_H}\left( {\dfrac{1}{{{4^2}}}} \right) $ (since $ \dfrac{1}{\infty } $ is zero)
Then
 $\Rightarrow \dfrac{1}{{{\lambda _s}}} = 1.0968 \times {10^7}\left( {\dfrac{1}{{16}}} \right) $ (Rydberg constant $ {R_H} = 1.0968 \times {10^7}{m^{ - 1}} $ )
By multiplying, and then inverting both sides to make $ {\lambda _s} $ subject of the formula we have
$\Rightarrow {\lambda _s} = 1.4587... \times {10^{ - 6}}m $
$\therefore {\lambda _s} = 1.46 \times {10^{ - 6}}m $.

Note
In reality, besides hydrogen, other elements also have their unique wavelength for each line of each series which also obeys the Rydberg formula (with some modification), and together these emission lines are called atomic spectra lines. In application, due to their uniqueness to different elements (i.e. no two elements have the same spectral lines), they can be used to identify unknown elements.
Alternatively, without first finding the Rydberg constant, we could just compare $ {\lambda _\alpha } $ and $ {\lambda _\gamma } $ by dividing, as in:
$\Rightarrow \dfrac{1}{{{\lambda _\alpha }}} \div \dfrac{1}{{{\lambda _\gamma }}} = {R_H}\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}} \right) \div \left[ {{R_H}\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{5^2}}}} \right)} \right] $
$\Rightarrow \dfrac{1}{{{\lambda _\alpha }}} \times \dfrac{{{\lambda _\gamma }}}{1} = {R_H}\left( {\dfrac{1}{4} - \dfrac{1}{9}} \right) \div \left[ {{R_H}\left( {\dfrac{1}{4} - \dfrac{1}{{25}}} \right)} \right] $
Evaluating the brackets we have
$\Rightarrow \dfrac{{{\lambda _\gamma }}}{{{\lambda _\alpha }}} = {R_H}\left( {\dfrac{5}{{36}}} \right) \div \left[ {{R_H}\left( {\dfrac{{21}}{{100}}} \right)} \right] $
$\Rightarrow \dfrac{{{\lambda _\gamma }}}{{{\lambda _\alpha }}} = \left( {\dfrac{{5{R_H}}}{{36}}} \right) \times \left( {\dfrac{{100}}{{21{R_H}}}} \right) $
Eliminating $ {R_H} $ , and evaluating right hand side, we have
 $\Rightarrow \dfrac{{{\lambda _\gamma }}}{{{\lambda _\alpha }}} = \dfrac{{125}}{{189}} $
Multiplying both sides by $ {\lambda _\alpha } $ and inserting its value, we have
$\Rightarrow {\lambda _\gamma } = \dfrac{{125}}{{189}} \times 6.563 \times {10^{ - 7}} $
$ \Rightarrow 4.340 \times {10^{ - 7}}m $.