What will be the ratio of wavelength of 3^rd line at Paschen series to 2^nd line of Balmer series of H-atom?

The correct option is 9/4.

To find the ratio of the wavelengths of the 3^rd line of the Paschen series to the 2^nd line of the Balmer series of a hydrogen atom, we use the Rydberg formula for the wavelength ($\mapsto$) of spectral lines:

$\frac{1}{\lambda }=R{Z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

For Hydrogen atom, atomic number $Z=1$.

Step 1: Calculate the wavelength for the 3^rd line of the Paschen series (${\lambda }_P$):
For the Paschen series, the lower energy level is $n_1=3$.
The 3^rd line corresponds to transitions from $n_2=n_1+3=3+3=6$.
Substituting these values into the Rydberg formula:

$\frac{1}{{\lambda }_P}=R\left(\frac{1}{3^2}-\frac{1}{6^2}\right)$

$\frac{1}{{\lambda }_P}=R\left(\frac{1}{9}-\frac{1}{36}\right)=R\left(\frac{4-1}{36}\right)=\frac{3R}{36}=\frac{R}{12}$

Therefore, the wavelength is:

${\lambda }_P=\frac{12}{R}$

Step 2: Calculate the wavelength for the 2^nd line of the Balmer series (${\lambda }_B$):
For the Balmer series, the lower energy level is $n_1=2$.
The 2^nd line corresponds to transitions from $n_2=n_1+2=2+2=4$.
Substituting these values into the Rydberg formula:

$\frac{1}{{\lambda }_B}=R\left(\frac{1}{2^2}-\frac{1}{4^2}\right)$

$\frac{1}{{\lambda }_B}=R\left(\frac{1}{4}-\frac{1}{16}\right)=R\left(\frac{4-1}{16}\right)=\frac{3R}{16}$

Therefore, the wavelength is:

${\lambda }_B=\frac{16}{3R}$

Step 3: Calculate the ratio of wavelengths $\frac{{\lambda }_P}{{\lambda }_B}$:

$\frac{{\lambda }_P}{{\lambda }_B}=\frac{\left(\frac{12}{R}\right)}{\left(\frac{16}{3R}\right)}=\frac{12}{R}\times \frac{3R}{16}=\frac{12\times 3}{16}=\frac{36}{16}=\frac{9}{4}$

Thus, the ratio of the wavelength of the 3^rd line of the Paschen series to the 2^nd line of the Balmer series is 9/4.