Question

What will be the ratio of the wavelength of the first line to that of the second line of the paschen series of H atoms?
\[(A)256:175\]
$(B)175:256$
$(C)15:16$
$(D)24:27$

Answer 1

Hint: This question is based on the concept of spectral series. Spectral series is a series of lines in the spectrum of light which is emitted by the excited atoms of an element. In spectral series, each line is related to the others in the series by a simple numerical equation.


Complete step by step answer:
Paschen series are the series of lines in the spectrum of the hydrogen atom which correspond to transitions between the state with principal quantum number n = 3 and successive higher states.
We know that the rydberg equation is,
$\dfrac{1}{\lambda } = R.\left( {\dfrac{1}{{n_f^2}} - \dfrac{1}{{n_i^2}}} \right)...........(1)$
Where,
$\lambda $ is the wavelength of the photon emitted by the electron.
$R$ is the rydberg constant which is equal to $1.097 \times {10^7}{m^{ - 1}}$
${n_f}$ is the final energy level of the electron
${n_i}$ is the initial energy level of the electron.
According to question,
The paschen series is characterized by ${n_f} = 3$
The first transition in the Paschen series corresponds to,
${n_i} = 4$ to ${n_f} = 3$
So, we can say that in this transition the electron drops from fourth energy level to third energy level,
On putting the values ${n_i} = 4$ and ${n_f} = 3$ in equation (1), we get,
$\dfrac{1}{{{\lambda _1}}} = R.\left( {\dfrac{1}{{{3^2}}} - \dfrac{1}{{{4^2}}}} \right)$
$\dfrac{1}{{{\lambda _1}}} = R.\left( {\dfrac{1}{9} - \dfrac{1}{{16}}} \right)$
On taking the LCM,
$\dfrac{1}{{{\lambda _1}}} = R.\left( {\dfrac{{16 - 9}}{{16 \times 9}}} \right)$
$\dfrac{1}{{{\lambda _1}}} = R.\left( {\dfrac{7}{{144}}} \right)........(2)$
Now, the second transition takes place from ${n_i} = 5$ to ${n_f} = 3$,
$\dfrac{1}{{{\lambda _2}}} = R.\left( {\dfrac{1}{{{3^2}}} - \dfrac{1}{{{5^2}}}} \right)$
$\dfrac{1}{{{\lambda _2}}} = R.\left( {\dfrac{1}{9} - \dfrac{1}{{25}}} \right)$
On taking the LCM,
$\dfrac{1}{{{\lambda _2}}} = R.\left( {\dfrac{{25 - 9}}{{25 \times 9}}} \right)$
$\dfrac{1}{{{\lambda _2}}} = R.\left( {\dfrac{{16}}{{225}}} \right)..........(3)$
On dividing equation (2) by (1),
$\dfrac{{\dfrac{1}{{{\lambda _2}}}}}{{\dfrac{1}{{{\lambda _1}}}}} = \dfrac{{\dfrac{{16R}}{{225}}}}{{\dfrac{{7R}}{{144}}}}$
On cancelling $R$,
$\dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{{144 \times 16}}{{225 \times 7}}$
$\dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{{2304}}{{1575}}$
On further simplifying, we get $\dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{{256}}{{175}}$
So, the final answer is \[(A)256:175\].

Note: Apart from the paschen spectral series on which this question was based, there are various other hydrogen spectral series. The names of these series are Lyman series, Balmer series, Brackett series, Pfund series, Humphreys series. Apart from these series, there are various other series which have been discovered but are yet to be named.