Question

Ratio of longest wavelengths corresponding to Lyman and Balmer series in hydrogen spectrum is
A. \[\dfrac{5}{27}\]
B. \[\dfrac{3}{23}\]
C. \[\dfrac{7}{29}\]
D. \[\dfrac{9}{31}\]

Answer 1

Hint: In this question we have been asked to calculate the ratio of longest wavelengths corresponding to Lyman and Balmer series in the hydrogen spectrum. Therefore, in order to solve this question, we shall first calculate the wavelengths using the Rydberg’s formula. Rydberg formula is an equation that can be used to determine the frequency, wavelength or energy of photons emitted.

Formula used:
\[\dfrac{1}{\lambda }=ZR\left[ \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right]\]
Where,
\[\lambda \] is the wavelength
Z is the atomic number
R is the Rydberg’s constant
\[{{n}_{1}},{{n}_{2}}\] are the principal quantum number of lower and higher energy levels respectively.

Complete step-by-step answer:
 We know that the emission spectrum of a hydrogen atom is divided into a number of spectral series. The wavelength of the same can be calculated using Rydberg formula.
Therefore, from Rydberg formula we know
\[\dfrac{1}{\lambda }=ZR\left[ \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right]\]
We know that in the spectral line for Lyman series is the transition from n = 2 to n= 1
Therefore, we can say wavelength for Lyman series,
\[\dfrac{1}{{{\lambda }_{L}}}=ZR\left[ \dfrac{1}{1}-\dfrac{1}{4} \right]\]
\[\dfrac{1}{{{\lambda }_{L}}}=\dfrac{4}{3ZR}\] …………… (1)
Similarly, for Balmer series the transition of spectral lines is from n = 3 to n = 2
Therefore,
\[\dfrac{1}{{{\lambda }_{B}}}=ZR\left[ \dfrac{1}{4}-\dfrac{1}{9} \right]\]
Therefore,
\[\dfrac{1}{{{\lambda }_{B}}}=\dfrac{36}{5ZR}\] …………… (2)
Now taking the ratio
From (1) and (2)
We get,
\[\dfrac{{{\lambda }_{L}}}{{{\lambda }_{B}}}=\dfrac{4}{3ZR}\times \dfrac{5ZR}{36}\]
Therefore,
\[\dfrac{{{\lambda }_{L}}}{{{\lambda }_{B}}}=\dfrac{5}{27}\]

So, the correct answer is “Option A”.

Note: The Lyman series was discovered by Theodore Lyman and the Balmer series was discovered by Johann Balmer. The equation by Rydberg can be applied to any system with a single particle orbiting a nucleus. The electron in a hydrogen atom experiences the transition when a hydrogen atom absorbs a photon. This causes the spectral formation in hydrogen atoms. All the spectral lines are due to the electrons moving in energy levels in an atom. The spectral series are important for calculating the red shifts.