Question 434: Atoms & Nuclei

Question

If ' $\lambda_1$ ' and ' $\lambda_2$ ' are the wavelengths of the first member of the Balmer and Paschen series, in hydrogen atom respectively, then the ratio of respective frequencies, $f_1 / f_2$, is

Accepted Answer

First, let us identify the transitions for the first members of the
Balmer
and
Paschen
series in the hydrogen atom:
Balmer series
(visible): transitions to $n_2 = 2$
First member: $n_1 = 3$ to $n_2 = 2$
Paschen series
(infrared): transitions to $n_2 = 3$
First member: $n_1 = 4$ to $n_2 = 3$
Let wavelengths be:
$\lambda_1$ = Balmer, $\displaystyle 3 ightarrow 2$
$\lambda_2$ = Paschen, $\displaystyle 4 ightarrow 3$
The
wavelength
for a transition in the hydrogen atom is given by the Rydberg formula:
$$ \frac{1}{\lambda} = R_H \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} ight) $$
For the
first Balmer line
$(3 	o 2)$:
$$ \frac{1}{\lambda_1} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R_H \left( \frac{1}{4} - \frac{1}{9} ight) $$
Calculate:
$$ \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} $$
So,
$$ \frac{1}{\lambda_1} = R_H \cdot \frac{5}{36} $$
For the
first Paschen line
$(4 	o 3)$:
$$ \frac{1}{\lambda_2} = R_H \left( \frac{1}{3^2} - \frac{1}{4^2} ight) = R_H \left( \frac{1}{9} - \frac{1}{16} ight) $$
Calculate:
$$ \frac{1}{9} - \frac{1}{16} = \frac{16 - 9}{144} = \frac{7}{144} $$
So,
$$ \frac{1}{\lambda_2} = R_H \cdot \frac{7}{144} $$
The
frequency
is given by $f = \dfrac{c}{\lambda}$
Therefore,
$$ f_1 = \frac{c}{\lambda_1}, \quad f_2 = \frac{c}{\lambda_2} $$
$$ \frac{f_1}{f_2} = \frac{\lambda_2}{\lambda_1} $$
From above:
$$ \frac{1}{\lambda_1} = R_H \cdot \frac{5}{36} \implies \lambda_1 = \frac{36}{5 R_H} $$
$$ \frac{1}{\lambda_2} = R_H \cdot \frac{7}{144} \implies \lambda_2 = \frac{144}{7 R_H} $$
Therefore,
$ \frac{f_1}{f_2} = \frac{\lambda_2}{\lambda_1} = \frac{\frac{144}{7 R_H}}{\frac{36}{5 R_H}} = \frac{144}{7} 	imes \frac{5}{36} $
$ = \frac{144 	imes 5}{7 	imes 36} $
$ = \frac{720}{252} $
$ = \frac{720 \div 36}{252 \div 36} = \frac{20}{7} $
Correct answer:
$ \boxed{20:7} $
So, the answer is
Option A
.

Answer

$20: 7$

Answer

$27: 5$

Answer

$50: 9$

Answer

$108: 7$

Atoms & Nuclei

If ' \lambda_1 ' and ' \lambda_2 ' are the wavelengths of the first member of the Balmer and Paschen series, in hydrogen atom respectively, then the ratio of respective frequencies, f_1 / f_2, is

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