
In the series in the line spectrum of hydrogen, the wavelength of radiation is $6.563 A^o$. The name of the series and the orbits in which electron transition takes place are:
A.Balmer series, 3rd to 2nd orbit
B.Lyman series, 2nd to 1st orbit
C.Pfund series, 6th to 5th orbit
D.Paschen series, 4th to 3rd orbit
Answer
591.6k+ views
Hint: We must know that the line spectrum of hydrogen is divided into several series with wavelengths given by the Rydberg formula. Put the values in the Rydberg formula to find the orbits.
Formula Used:
Rydberg formula is given below as,
\[\dfrac{1}{\lambda } = {R_H}[\dfrac{1}{{{{({n_1})}^2}}} - \dfrac{{1}}{{{{({n_2})}^2}}}]\],
Where,
$n_1$ and $n_2$ are the orbits and
$R_H$ is the Rydberg constant (Rydberg constant is equal to $1.09 \times 10^{-7}m$).
Complete step by step answer:Let’s start with discussing a little about the line spectrum of hydrogen for better understanding of the question. The line spectrum of hydrogen is divided into several series with wavelengths given by the Rydberg formula. The Rydberg formula is given down below as,
\[\dfrac{1}{\lambda } = {R_H}[\dfrac{1}{{{{({n_1})}^2}}} - \dfrac{{1}}{{{{({n_2})}^2}}}]\],
Where,
$n_1$ and $n_2$ are the orbits,
In the question we are given the wavelength of radiation which is $6.563 A^o$.
Putting this value into the formula we get,
$ \Rightarrow $\[\dfrac{1}{{6.563{\text{ }} \times {\text{ 1}}{{\text{0}}^{ - 7}}m}}{\text{ }} = {\text{ 1}}{\text{.09 }} \times {\text{ 1}}{{\text{0}}^7}[\dfrac{1}{{{{({n_1})}^2}}} - \dfrac{{1}}{{{{({n_2})}^2}}}]\]
Solving this we get,
$ \Rightarrow $$\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}} = \dfrac{5}{{36}}$
Which gives $n_1 = 2$ and $n_2 = 3$.
Also the series between the 2nd and 3rd orbit is the Balmer series.
Hence the answer to this question is option A. Balmer series, 3rd to 2nd orbit.
Note:We can remember that the line spectrum of hydrogen mainly consist of 6 series which are Lyman series (from orbit 1), Balmer series (From orbit 2), Paschen series which is also known as Bohr series (from orbit 3), Brackett Series (from orbit 4), Pfund Series (from orbit 5) and Humphreys series (from orbit 6). There are series above this as well but they are not used that much.
Formula Used:
Rydberg formula is given below as,
\[\dfrac{1}{\lambda } = {R_H}[\dfrac{1}{{{{({n_1})}^2}}} - \dfrac{{1}}{{{{({n_2})}^2}}}]\],
Where,
$n_1$ and $n_2$ are the orbits and
$R_H$ is the Rydberg constant (Rydberg constant is equal to $1.09 \times 10^{-7}m$).
Complete step by step answer:Let’s start with discussing a little about the line spectrum of hydrogen for better understanding of the question. The line spectrum of hydrogen is divided into several series with wavelengths given by the Rydberg formula. The Rydberg formula is given down below as,
\[\dfrac{1}{\lambda } = {R_H}[\dfrac{1}{{{{({n_1})}^2}}} - \dfrac{{1}}{{{{({n_2})}^2}}}]\],
Where,
$n_1$ and $n_2$ are the orbits,
In the question we are given the wavelength of radiation which is $6.563 A^o$.
Putting this value into the formula we get,
$ \Rightarrow $\[\dfrac{1}{{6.563{\text{ }} \times {\text{ 1}}{{\text{0}}^{ - 7}}m}}{\text{ }} = {\text{ 1}}{\text{.09 }} \times {\text{ 1}}{{\text{0}}^7}[\dfrac{1}{{{{({n_1})}^2}}} - \dfrac{{1}}{{{{({n_2})}^2}}}]\]
Solving this we get,
$ \Rightarrow $$\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}} = \dfrac{5}{{36}}$
Which gives $n_1 = 2$ and $n_2 = 3$.
Also the series between the 2nd and 3rd orbit is the Balmer series.
Hence the answer to this question is option A. Balmer series, 3rd to 2nd orbit.
Note:We can remember that the line spectrum of hydrogen mainly consist of 6 series which are Lyman series (from orbit 1), Balmer series (From orbit 2), Paschen series which is also known as Bohr series (from orbit 3), Brackett Series (from orbit 4), Pfund Series (from orbit 5) and Humphreys series (from orbit 6). There are series above this as well but they are not used that much.
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