Which of the following spectral series of hydrogen atoms is lying in a visible range of electromagnetic waves?
(A) Paschen series
(B) Pfund series
(C) Lyman series
(D) Balmer series
Answer
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Hint To solve this question, we need to find out the range of the wavelength of the spectral lines corresponding to each series. We have to apply the formula representing each series, given as $ \dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right) $ , for determining the range. And then, using the range for the visible region of the electromagnetic spectrum we can get the final answer.
Formula used: The formula which is used to solve this question is given by
$ \dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right) $ , here $ \lambda $ is the wavelength of the spectral line, and $ R $ is the Rydberg’s constant.
Complete step by step solution:
We know that the Paschen series is represented by the formula
$ \dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{3^2}}} - \dfrac{1}{{{n^2}}}} \right) $ …………..(1)
For the maximum wavelength corresponding to this series, we substitute $ n = 4 $ in the above equation. So we get
$\Rightarrow \dfrac{1}{{{\lambda _L}}} = R\left( {\dfrac{1}{{{3^2}}} - \dfrac{1}{{{4^2}}}} \right) $
On simplifying and taking the reciprocal, we get
$\Rightarrow {\lambda _L} = \dfrac{{144}}{{7R}} $
We know that the value of the Rydberg constant is $ R = 1.096 \times {10^7}{m^{ - 1}} $ . So we get
$\Rightarrow {\lambda _L} = 1876.9nm $
For the minimum wavelength, we take the limit $ n \to \infty $ in (1). On taking the limit, we get
$\Rightarrow \dfrac{1}{{{\lambda _H}}} = R\left( {\dfrac{1}{{{3^2}}} - 0} \right) $
$\Rightarrow {\lambda _H} = \dfrac{9}{R} $
On substituting $ R = 1.096 \times {10^7}{m^{ - 1}} $ , we get
$\Rightarrow {\lambda _H} = 821.2nm $
So the range of the wavelengths corresponding to the Paschen series is from $ 821.2nm $ to $ 1876.9nm $ .
Similarly, we get the corresponding range for the Lyman series from $ 91nm $ to $ 122nm $ , and for the Balmer series from $ 365nm $ to $ 656nm $ .
We know that the visible range of electromagnetic waves is from $ 400nm $ to $ 700nm $ , which clearly falls in the range of the Balmer series.
Hence, the correct answer is option D.
Note:
As we can see in the above solution that there are a lot of calculations involved for determining the range of the wavelengths. So, it is preferred to remember the type of electromagnetic radiation corresponding to each series to get the answer quickly.
Formula used: The formula which is used to solve this question is given by
$ \dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right) $ , here $ \lambda $ is the wavelength of the spectral line, and $ R $ is the Rydberg’s constant.
Complete step by step solution:
We know that the Paschen series is represented by the formula
$ \dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{3^2}}} - \dfrac{1}{{{n^2}}}} \right) $ …………..(1)
For the maximum wavelength corresponding to this series, we substitute $ n = 4 $ in the above equation. So we get
$\Rightarrow \dfrac{1}{{{\lambda _L}}} = R\left( {\dfrac{1}{{{3^2}}} - \dfrac{1}{{{4^2}}}} \right) $
On simplifying and taking the reciprocal, we get
$\Rightarrow {\lambda _L} = \dfrac{{144}}{{7R}} $
We know that the value of the Rydberg constant is $ R = 1.096 \times {10^7}{m^{ - 1}} $ . So we get
$\Rightarrow {\lambda _L} = 1876.9nm $
For the minimum wavelength, we take the limit $ n \to \infty $ in (1). On taking the limit, we get
$\Rightarrow \dfrac{1}{{{\lambda _H}}} = R\left( {\dfrac{1}{{{3^2}}} - 0} \right) $
$\Rightarrow {\lambda _H} = \dfrac{9}{R} $
On substituting $ R = 1.096 \times {10^7}{m^{ - 1}} $ , we get
$\Rightarrow {\lambda _H} = 821.2nm $
So the range of the wavelengths corresponding to the Paschen series is from $ 821.2nm $ to $ 1876.9nm $ .
Similarly, we get the corresponding range for the Lyman series from $ 91nm $ to $ 122nm $ , and for the Balmer series from $ 365nm $ to $ 656nm $ .
We know that the visible range of electromagnetic waves is from $ 400nm $ to $ 700nm $ , which clearly falls in the range of the Balmer series.
Hence, the correct answer is option D.
Note:
As we can see in the above solution that there are a lot of calculations involved for determining the range of the wavelengths. So, it is preferred to remember the type of electromagnetic radiation corresponding to each series to get the answer quickly.
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