Question

What would be the maximum wavelength for Bracket series of hydrogen spectrum?
A. $74483{{A}^{{}^\circ }}$
B. $22790{{A}^{{}^\circ }}$
C. $40519{{A}^{{}^\circ }}$
D. $18753{{A}^{{}^\circ }}$

Answer 1

You could recall the expression for wavelength for the radiation emitted or you could derive it based on the Bohr’s postulate. We are asked the maximum wavelength, so, remember that the transition of electrons between the adjacent orbits will have small energy differences. Otherwise, we could say that, frequency emitted will be small for such a transition hence, maximum will be the wavelength. Also, $R=1.09687\times {{10}^{7}}{{m}^{-1}}$

Formula used:
Expression for wavelength based on Bohr’s postulate,
$\dfrac{1}{\lambda }=R\left( \dfrac{1}{{{n}_{f}}^{2}}-\dfrac{1}{{{n}_{i}}^{2}} \right)$

Complete answer:
It was expected of the frequencies of light emitted by a particular element to show a regular pattern, also, hydrogen being the simplest atom should exhibit the simplest pattern. The spacing between lines within certain sets of the hydrogen spectrum does decrease in a regular way and these sets are what we call the spectral lines.
Let us recall the third postulate of Bohr’s model which states that, as long as the electron is present in one of the stationary orbits, it doesn’t radiate energy. However, when a transition from higher stationary orbit to lower stationary orbit happens for an electron, it does radiate energy that is equal to the energy difference of the electron in the two orbits.
Now let us derive Bohr’s formula for spectral lines in hydrogen spectrum,
Let, ${{E}_{i}}$ be the energy of electron in the initial ${{n}_{i}}th$ orbit and let ${{E}_{f}}$ be the energy of the electron in the final ${{n}_{f}}th$ orbit, then, by Bohr’s third postulate,
${{E}_{i}}-{{E}_{f}}=h\nu$
$\Rightarrow \nu =\dfrac{{{E}_{i}}-{{E}_{f}}}{h}$ ………………………… (1)
We know that the total energy of an electron in the stationary states of hydrogen atom is given by,
${{E}_{n}}=-\dfrac{m{{e}^{4}}}{8{{\varepsilon }_{0}}{{h}^{2}}{{n}^{2}}}$
$\Rightarrow {{E}_{i}}=-\dfrac{m{{e}^{4}}}{8{{\varepsilon }_{0}}{{h}^{2}}{{n}_{i}}^{2}}$ ……………………………. (2)
$\Rightarrow {{E}_{f}}=-\dfrac{m{{e}^{4}}}{8{{\varepsilon }_{0}}{{h}^{2}}{{n}_{f}}^{2}}$………………………………… (3)
Substituting (3) and (2) in (1), we get,
$\Rightarrow \nu =\dfrac{\left( -\dfrac{m{{e}^{4}}}{8{{\varepsilon }_{0}}{{h}^{2}}{{n}_{i}}^{2}} \right)-\left( -\dfrac{m{{e}^{4}}}{8{{\varepsilon }_{0}}{{h}^{2}}{{n}_{f}}^{2}} \right)}{h}$
$\Rightarrow \nu =\dfrac{m{{e}^{4}}}{8{{\varepsilon }_{0}}^{2}{{h}^{3}}}\left( -\dfrac{1}{{{n}_{i}}^{2}}+\dfrac{1}{{{n}_{f}}^{2}} \right)$
But we have,
$\nu =\dfrac{c}{\lambda }$
Where, c is the speed of electromagnetic radiation.
$\Rightarrow \dfrac{c}{\lambda }=\dfrac{m{{e}^{4}}}{8{{\varepsilon }_{0}}^{2}{{h}^{3}}}\left( \dfrac{1}{{{n}_{f}}^{2}}-\dfrac{1}{{{n}_{i}}^{2}} \right)$
$\Rightarrow \dfrac{1}{\lambda }=R\left( \dfrac{1}{{{n}_{f}}^{2}}-\dfrac{1}{{{n}_{i}}^{2}} \right)$ ……………………………. (4)
Where, R is the Rydberg’s constant given by,
$R=\dfrac{m{{e}^{4}}}{8{{\varepsilon }_{0}}^{2}{{h}^{3}}c}$
We know that the bracket series is one among the series of spectra of hydrogen and others include Balmer, Lyman, Paschen and Pfund. Bracket series are found in the Infrared region.
Equation (4) in case of Brackett series can be given by,
$\Rightarrow \dfrac{1}{\lambda }=R\left( \dfrac{1}{{{4}^{2}}}-\dfrac{1}{{{n}^{2}}} \right)$………………………………. (5)
Where, n=5, 6, 7…
Since we are asked to find the maximum wavelength for Bracket series,
$n=5$
Now (5) becomes,
$\Rightarrow \dfrac{1}{\lambda }=R\left( \dfrac{1}{{{4}^{2}}}-\dfrac{1}{{{5}^{2}}} \right)$
But we know,$R=1.09687\times {{10}^{7}}{{m}^{-1}}$
$\Rightarrow \dfrac{1}{\lambda }=1.09687\times {{10}^{7}}{{m}^{-1}}\left( \dfrac{1}{{{4}^{2}}}-\dfrac{1}{{{5}^{2}}} \right)$
$\Rightarrow \lambda =\dfrac{400}{9\times 1.09687\times {{10}^{7}}}m$
$\Rightarrow \lambda =40.519\times {{10}^{-7}}m$
$\Rightarrow \lambda =40519{{A}^{{}^\circ }}$
Therefore, maximum wavelength for Bracket series of hydrogen spectrum will be, $\lambda =40519{{A}^{{}^\circ }}$

Hence, the answer to the given question is option C.

Note:
We know that, the wave number is given by,
$\overline{\nu }=\dfrac{1}{\lambda }$
So equation (4) can be rewritten as,
$\Rightarrow \overline{\nu }=R\left( \dfrac{1}{{{n}_{f}}^{2}}-\dfrac{1}{{{n}_{i}}^{2}} \right)$
We have substituted n=5 (nearest orbit) for maximum wavelength due to the obvious reason that the energy difference between the two adjacent orbits would be less, therefore the frequency of the radiation will be low and hence wavelength will be maximum. Similarly, for minimum wavelength we substitute $n=\infty$.