Question

The shortest wavelength Of ${\mathrm{He}}^{+}$ ion in the Balmer Series is x, then the longest wavelength in the Paschen Series Of ${\mathrm{Li}}^{+2}$ Is:

Answer 1

Step 1: Finding the shortest wavelength in${\mathrm{He}}^{+}$, the Balmer series:

• The Rydberg equation is :$\frac{1}{\mathrm{\lambda }}=R_H\left[Z^2\right]\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$
• where $Z=$Atomic no. , ${\mathrm{n}}_{1,}{\mathrm{n}}_2=$Energy levels wherein ${\mathrm{n}}_1$is always greater than ${\mathrm{n}}_2$, and $\mathrm{R}=$Rydberg constant.
• For finding the shortest wavelength in the Balmer series${\mathrm{n}}_1$$=2$, ${\mathrm{n}}_2$$=\infty$ and $Z$ of ${\mathrm{He}}^{+}$is 2.
• So, $\frac{1}{\mathrm{\lambda }}=R\left[2^2\right]\left[\frac{1}{2^2}-\frac{1}{\infty }\right]$
• Where, $\mathrm{\lambda }=\mathrm{x}$[given]
• By solving, $\frac{1}{\mathrm{x}}=R$
• and, $\mathrm{x}=\frac{1}{\mathrm{R}}$

Step 2: Finding the shortest wavelength in${\mathrm{Li}}^{+2}$, the Paschen series:

• The Rydberg equation is :$\frac{1}{\mathrm{\lambda }}=R_H\left[Z^2\right]\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$
• where $Z=$Atomic no. , ${\mathrm{n}}_{1,}{\mathrm{n}}_2=$Energy levels wherein ${\mathrm{n}}_1$is always greater than ${\mathrm{n}}_2$, and $\mathrm{R}=$Rydberg constant.
• For finding the longest wavelength of the Paschen series${\mathrm{n}}_1$$=3$, ${\mathrm{n}}_2$$=4$and $Z$ of ${\mathrm{Li}}^{+2}$ is $3$.
• So,$\frac{1}{\lambda }=R\left[3^2\right]\left[\frac{1}{3^2}-\frac{1}{4^2}\right]$
• By solving, $\frac{1}{\mathrm{\lambda }}=R\times \frac{7}{16}$

Step 3: Dividing$\mathrm{\lambda }$ by$\mathrm{x}$.

• where, $\frac{1}{\mathrm{\lambda }}=R\times \frac{7}{16}$
• and, $\mathrm{x}=\frac{1}{\mathrm{R}}$
• then, $\frac{\mathrm{\lambda }}{\mathrm{x}}=\frac{16}{7R}\times \frac{\mathrm{R}}{1}$
• By solving, $\mathrm{\lambda }=\frac{16}{7}\mathrm{x}$

Hence, In ${\mathrm{Li}}^{+2}$the longest wavelength in the Paschen series is $\mathrm{\lambda }=\frac{16}{7}\mathrm{x}$.