Question

If the series limit frequency of the Lyman series is ${\nu _L}$, then the series limit frequency of the PFund series is?
a) $\dfrac{{{\nu _L}}}{{16}}$
b) $\dfrac{{{\nu _L}}}{{25}}$
c) $25{\nu _L}$
d) $16{\nu _L}$

Answer 1

Compare the Lyman series and PFund series with the help of Rydberg formula that describes wavelength. Further calculation on the wavelength will help to solve this problem.

Formula Used:
$\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right)$
Where, $\lambda =$Wavelength, $R =$ Rydberg constant

Complete step by step answer:
Let’s discuss the Lyman series first; this series is a related sequence of wavelengths that describe electromagnetic energy given off by energized atoms in the ultraviolet region. When coming to the PFund series, it is a series of lines which lies in the infrared part of the spectrum of atomic hydrogen
We can solve this problem by using the Rydberg formula. That is,
$\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right)$
Where, $\lambda =$ Wavelength
$R =$ Rydberg constant
In Lyman series,
Value of ${n_1} = 1$
And the value of ${n_2} = \infty$
For Lyman series, assume the value of $\lambda$ as, $\lambda = {\lambda _L}$
Applying the values to the Rydberg formula, we get,
$\dfrac{1}{{{\lambda _L}}} = R\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{\infty ^2}}}} \right)$
When any quantity is divided by infinity the value of the resultant is nearly zero.
$\Rightarrow \dfrac{1}{{{\lambda _L}}} = R$………………………………………………… (Eqn. A)

In PFund series,
Value of ${n_1} = 5$
And the value of ${n_2} = \infty$
For $P$Fund series, assume the value of $\lambda$ as, $\lambda = {\lambda _P}$
Applying the values to the Rydberg formula, we get,
$\dfrac{1}{{{\lambda _P}}} = R\left( {\dfrac{1}{{{5^2}}} - \dfrac{1}{{{\infty ^2}}}} \right)$
As we discussed earlier, when any quantity is divided by infinity the value of the resultant is nearly zero.
$\Rightarrow \dfrac{1}{{{\lambda _P}}} = \dfrac{R}{{25}}$………………………………………………… (Eqn. B)
Comparing Eqn. A and Eqn. B, we will get,
$\Rightarrow \dfrac{1}{{{\lambda _P}}} = \dfrac{1}{{25{\lambda _L}}}$………………………………………….. (Eqn. C)
We know, $\nu = \dfrac{c}{\lambda }$
$\Rightarrow \lambda = \dfrac{c}{\nu }$
Applying this value in the Equation C, we get,
$\Rightarrow \dfrac{{{\nu _P}}}{c} = \dfrac{{{\nu _L}}}{{25c}}$
$\Rightarrow {\nu _P} = \dfrac{{{\nu _L}}}{{25}}$
So the final answer is Option (B)

Note: Origin of the PFund series is in the 1930s; it is a series of absorption or emission lines of atomic hydrogen. The lines were experimentally discovered in 1924 by August Herman Pfund, and correspond to the electron jumping between the fifth and higher energy levels of the hydrogen atom.