Question

In the spectrum of hydrogen, the ratio of the longest wavelength in Lyman series to the longest wavelength in Balmer series is:
A. $\dfrac{5}{27}$
B. $\dfrac{4}{9}$
C. $\dfrac{9}{4}$
D. $\dfrac{27}{5}$

Answer 1

Hint: Electron jumps from higher orbital to 1st orbital in Lyman series and second orbital in Balmer series. Energy emitted for transitions of electrons can be expressed as $E=\dfrac{hc}{\lambda }$. For the longest wavelength, energy released should be minimum.

Formula used:
$\dfrac{1}{\lambda }={{R}_{H}}{{z}^{2}}\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right)$

Complete step-by-step answer:
Lyman series and Balmer series are special parts of the spectrum of hydrogen with definite starting and final orbital through which the electronic transition occurs.
Now, for electronic transitions between two orbits, when an electron jumps from higher orbital to lower orbital, energy is released in the form of spectrum.
Lyman series correspond to electronic transitions from higher orbital to the innermost orbital, while Balmar series correspond to transitions from higher orbital to second innermost orbital, whatever be the orbital from which electron jumps.
Now, for such electronic transitions, we have a formula for calculating the wavelength of spectrum when an electron jumps from higher to lower orbital, i.e.
$\dfrac{1}{\lambda }={{R}_{H}}{{z}^{2}}\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right)$
Where, ${{n}_{1}}$ corresponds to the final orbital in which electron lands,
${{n}_{2}}$ corresponds to the initial orbital from which electron jumps,
${{R}_{H}}$is the Rydberg constant,
And z is the atomic number of element in which transition occurs
Thus, for hydrogen atom, $z=1$
For Lyman series: ${{n}_{1}}=1$ and ${{n}_{2}}=2,3,4...$
For Balmer series: ${{n}_{1}}=2$ and ${{n}_{2}}=3,4,5...$
Now, for computing maximum $\lambda$by the above formula, for fixed ${{n}_{1}}$ :
$\dfrac{1}{\lambda }={{R}_{H}}\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right)$
i. ${{n}_{2}}$ should be minimum
ii. For maximum $\lambda$ of Lyman series: ${{n}_{1}}=1$ and ${{n}_{2}}=2$
iii. For maximum $\lambda$ of Balmer series: ${{n}_{1}}=2$ and ${{n}_{2}}=3$
Thus,
For Lyman series-
$\dfrac{1}{{{\lambda }_{\max }}}={{R}_{H}}\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right)={{R}_{H}}\left( \dfrac{1}{1}-\dfrac{1}{4} \right)=\dfrac{3{{R}_{H}}}{4}$
${{\lambda }_{\max }}\left( L \right)=\dfrac{4}{3{{R}_{H}}}$
For Balmer series-
$\dfrac{1}{{{\lambda }_{\max }}}={{R}_{H}}\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right)={{R}_{H}}\left( \dfrac{1}{4}-\dfrac{1}{9} \right)=\dfrac{5{{R}_{H}}}{36}$
${{\lambda }_{\max }}\left( B \right)=\dfrac{36}{5{{R}_{H}}}$
Thus, ratio of wavelengths-
$\dfrac{{{\lambda }_{\max }}\left( L \right)}{{{\lambda }_{\max }}\left( B \right)}=\dfrac{\dfrac{4}{3{{R}_{H}}}}{\dfrac{36}{5{{R}_{H}}}}=\dfrac{4\times 5{{R}_{H}}}{36\times 3{{R}_{H}}}=\dfrac{5}{27}$
The ratio of the longest wavelength in Lyman series to the longest wavelength in Balmer series is $\dfrac{5}{27}$
Hence, the correct option is A.

Note: Students should remember the definition of Lyman series and Balmer series and should not get confused with orbital numbers related to transition in Lyman and Balmer series.
While considering maximum and minimum wavelengths in Lyman series and Balmer series, estimation of final orbital should be done with taking care of signs and relation in the formula used.