Question

Let ${v_1}$ be the frequency of series limit of Lyman series, ${v_2}$ the frequency of the first line of Lyman series, and ${v_3}$ the frequency of series limit of Balmer series. Then which of the following is correct?
A.${v_1} - {v_2} = {v_3}$
B.${v_2} - {v_1} = {v_3}$
C.${v_3} = \dfrac{1}{2}\left( {{v_1} + {v_2}} \right)$
D.${v_2} + {v_1} = {v_3}$

Answer 1

Hint: In this question, we will simply find out the Series limit for Balmer series and series limit for Lyman series. Then we will calculate the frequency and name them as ${\upsilon _1}$, ${\upsilon _2}$ and ${\upsilon _3}$ respectively. Refer to the solution below.

Complete answer:
Electron transmission for the ultimate line of Lyman series.
$ \Rightarrow {n_1} = \infty \to {n_2} = 1$
Thus, the frequency associated with this will be named as ${\upsilon _1}$. Which will be calculated as-
$
   \Rightarrow {\upsilon _1} = {R_H} \times C\left[ {\dfrac{1}{{{n_2}^2}} - \dfrac{1}{{{n_1}^2}}} \right] \\
    \\
   \Rightarrow {\upsilon _1} = {R_H} \times C\left[ {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{\infty ^2}}}} \right] \\
    \\
   \Rightarrow {\upsilon _1} = {R_H} \times C \\
$
Let’s consider ${R_H} \times C$ as R. And we will consider this equation as equation 1.
$ \Rightarrow {\upsilon _1} = R$ (equation 1)
Electron transmission for the first line of Lyman series.
$ \Rightarrow {n_1} = 2 \to {n_2} = 1$
Thus, the frequency associated with this will be named as ${\upsilon _2}$. Which will be calculated as-
$
   \Rightarrow {\upsilon _2} = {R_H} \times C\left[ {\dfrac{1}{{{n_2}^2}} - \dfrac{1}{{{n_1}^2}}} \right] \\
    \\
   \Rightarrow {\upsilon _2} = {R_H} \times C\left[ {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}} \right] \\
    \\
   \Rightarrow {\upsilon _2} = {R_H} \times C\left[ {\dfrac{3}{4}} \right] \\
$
Putting R instead of ${R_H} \times C$, we will get-
$ \Rightarrow {\upsilon _2} = \dfrac{3}{4}R$
Considering this equation as equation 2.
$ \Rightarrow {\upsilon _1} = R$ (equation 2)
Series limit for Balmer series when transmission of electrons will occur-
$ \Rightarrow {n_1} = \infty \to {n_2} = 2$
Thus, the frequency associated with this will be named as ${\upsilon _3}$. Which will be calculated as-
$
   \Rightarrow {\upsilon _3} = {R_H} \times C\left[ {\dfrac{1}{{{n_2}^2}} - \dfrac{1}{{{n_1}^2}}} \right] \\
    \\
   \Rightarrow {\upsilon _3} = {R_H} \times C\left[ {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{\infty ^2}}}} \right] \\
    \\
   \Rightarrow {\upsilon _3} = {R_H} \times C\left[ {\dfrac{1}{4}} \right] \\
$
Putting R instead of ${R_H} \times C$, we will get-
$ \Rightarrow {\upsilon _3} = \dfrac{1}{4}R$
Considering this equation as equation 3.
$ \Rightarrow {\upsilon _1} = R$ (equation 3)
Now, putting these values into the equations given in the options, we will have-
$
   \Rightarrow {\upsilon _1} - {\upsilon _2} = {\upsilon _3} \\
    \\
   \Rightarrow R - \dfrac{{3R}}{4} = \dfrac{R}{4} \\
    \\
   \Rightarrow \dfrac{{4R - 3R}}{4} = \dfrac{R}{4} \\
    \\
   \Rightarrow \dfrac{R}{4} = \dfrac{R}{4} \\
$
Hence, option A will be considered as the correct option.

Note: The Lyman series is a related wavelength sequence describing electro-magnetic energy produced in ultraviolet energy by energized atoms. The Balmer series is one of six designated series describing the hydrogen spectral emissions line of the atom. Balmer lines are used in atomic physics. The Balmer series is computed with the Balmer formula, an empirical equation which Johann Balmer discovered in 1885.