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**Hint:**By calculating the value of wavelength of each of the given options using the wavelength formula of hydrogen spectral lines i.e. $\dfrac{1}{\lambda }=R{{Z}^{2}}\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right)$ Here, R is the Rydberg constant , Z is the atomic number , ${{n}_{1}}$ corresponds to 1, 2, 3, 4 and 5 and ${{n}_{2}}$ tells from which spectral line the transition has taken place. In this we can easily find the series which has the same value of wavelength as that given in the statement.

**Complete step by step answer:**

First let’s discuss the line spectra of hydrogen . when an electric discharge is passed through gaseous hydrogen, the hydrogen molecule dissociates and the energetically excited hydrogen atoms produced emit electromagnetic radiation of discrete frequencies and the hydrogen spectrum consists of several series of lines. These are Lyman, Balmer , Paschen , Bracket and Pfund series.

We can calculate the wavelength for the spectral line of H-atom as;

$\dfrac{1}{\lambda }=R{{Z}^{2}}\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right)$

Here, R is called as the Rydberg constant , Z is the atomic number , ${{n}_{1}}$ corresponds to 1, 2, 3, 4 and 5 known as Lyman, Balmer , Paschen , Bracket and Pfund series and ${{n}_{2}}$ tells from which spectral line the transition has taken place.

Now considering the numerical;

We can know that which series will give wavelength of 900 nm by finding the value of wavelength for each option as;

(a) Using the above formula of wavelength as;

$\dfrac{1}{\lambda }=R{{Z}^{2}}\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right)$ --------(1)

For paschen series; ${{n}_{1}}=3$

${{n}_{2}}=5$

${{R}_{H}}=1\times {{10}^{5}}c{{m}^{-1}}$

=$1\times {{10}^{7}}{{m}^{-1}}$( 1m=100cm)

Z for hydrogen=1

Put these values in equation(1) ,we get;

$\dfrac{1}{\lambda }=1\times 1{{0}^{-7}}\times 1\left( \dfrac{1}{9}-\dfrac{1}{25} \right)$

=$1\times 1{{0}^{-7}}\times \left( \dfrac{25-9}{225} \right)$

$\dfrac{1}{\lambda }$ =$1\times 1{{0}^{-7}}\times \left( \dfrac{16}{225} \right)$

$\lambda $ =$1{{0}^{-7}}\times \dfrac{225}{16}$

= $14.06\times 1{{0}^{-7}}$m

= 1406 nm ( 1m= ${{10}^{9}}$nm)

As the value of wavelength is not the same , therefore, this option is incorrect.

(b) For paschen series; ${{n}_{1}}=3$

${{n}_{2}}=\infty $

${{R}_{H}}=1\times {{10}^{5}}c{{m}^{-1}}$

=$1\times {{10}^{7}}{{m}^{-1}}$( 1m=100cm)

Z for hydrogen=1

Put these values in equation(1) ,we get;

$\dfrac{1}{\lambda }=1\times 1{{0}^{-7}}\times 1\left( \dfrac{1}{9}-\dfrac{1}{\infty } \right)$

$\dfrac{1}{\lambda }$ =$1{{0}^{-7}}\times \dfrac{1}{9}$

$\lambda $ =$1{{0}^{-7}}\times 9$ m

= 900 nm ( 1m= ${{10}^{9}}$nm)

As the value of wavelength is the same , therefore, this option is correct.

(c) For Lyman series; ${{n}_{1}}=1$

${{n}_{2}}=\infty $

${{R}_{H}}=1\times {{10}^{5}}c{{m}^{-1}}$

=$1\times {{10}^{7}}{{m}^{-1}}$( 1m=100cm)

Z for hydrogen=1

Put these values in equation(1) ,we get;

$\dfrac{1}{\lambda }=1\times 1{{0}^{-7}}\times 1\left( \dfrac{1}{1}-\dfrac{1}{\infty } \right)$

$\dfrac{1}{\lambda }$ =$1{{0}^{-7}}\times \dfrac{1}{1}$

$\lambda $ =$\dfrac{1}{{{10}^{7}}}$

= 100 nm ( 1m= ${{10}^{9}}$nm)

As the value of wavelength is not the same , therefore, this option is incorrect.

(d) For Balmer, series; ${{n}_{1}}=2$

${{n}_{2}}=\infty $

${{R}_{H}}=1\times {{10}^{5}}c{{m}^{-1}}$

=$1\times {{10}^{7}}{{m}^{-1}}$( 1m=100cm)

Z for hydrogen=1

Put these values in equation(1) ,we get;

$\dfrac{1}{\lambda }=1\times 1{{0}^{-7}}\times 1\left( \dfrac{1}{4}-\dfrac{1}{\infty } \right)$

$\dfrac{1}{\lambda }$ =$1{{0}^{-7}}\times \dfrac{1}{4}$

$\lambda $ =$1{{0}^{-7}}\times 4$m

= 400 nm ( 1m= ${{10}^{9}}$nm)

As the value of wavelength is not the same , therefore, this option is incorrect.

Hence, for the heat treatment of muscular pain , the Paschen , $\infty \to 3$ spectral line of H-atom produces, the radiation of wavelength of about 900 nm is suitable for this purpose

So, option (b) is correct.

**Note:**the spectral lines of hydrogen atom i.e. the Lyman series lies in the ultraviolet region , Balmer series lies in the visible region , Paschen ,Brackett and Pfund series lies in the infrared regions.

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