Question

In the line spectra of hydrogen atoms, the difference between the largest and the shortest wavelengths of the Lyman series is $304\AA$. The corresponding difference for the Paschen series in Å is: ___________.

Answer 1

Step 1: Given

Difference between the largest and the shortest wavelengths of the Lyman series is $304\AA$.

Step 2: Use the formula shortest wavelength in Lyman, we have

$\lambda =\frac{c}{\left(\frac{1}{n_2^2-n_2^2}\right)}$ $\mathrm{\lambda }\mathrm{is}\mathrm{wavelength},\mathrm{C}is\mathrm{wavelength},{\mathrm{n}}_1\mathrm{lower}\mathrm{energy}\mathrm{state},{\mathrm{n}}_2\mathrm{higher}\mathrm{energy}\mathrm{state}$

Step 3: For the Lyman series

For Largest wavelength the highest energy state would be infinity and lowest would be 1.

${\lambda }_1=\frac{c}{\left(\frac{1}{1^2-{\infty }^2}\right)}=c[\because n=\infty \text{\hspace{0.17em}to}n=1]$

For samllest wavelength the highest energy state would be from 2 and lowest would be 1.

${\lambda }_2=\frac{c}{\frac{1}{1^2}-\frac{1}{2^2}}=\frac{4c}{3}[\because n=2\text{to}n=1]$

Now difference between wavelength is

$\begin{array}{l}\Delta \lambda ={\lambda }_2-{\lambda }_1=304A°\\ \therefore \frac{4c}{3}-c=304A°\\ \Rightarrow \frac{3}{c}=304A°\\ \Rightarrow c=912A°\end{array}$

Step 4: For the Paschen series:

For Largest wavelength the highest energy state would be infinity and lowest would be 3.

${\lambda }_1=\frac{c}{\left(\frac{1}{3^2-{\infty }^2}\right)}=9c[\because n=\infty \text{\hspace{0.17em}to}n=3]$

For samllest wavelength the highest energy state would be from 4 and lowest would be 3.

$\begin{array}{l}\\ {\lambda }_2=\frac{c}{\frac{1}{3^2}-\frac{1}{4^2}}=\frac{144c}{7}[\because n=4\text{to}n=3]\end{array}$

Find the difference

$\begin{array}{c}\Delta \lambda ={\lambda }_2-{\lambda }_1\\ =\frac{144c}{7}-9c\\ =\frac{81}{7}\\ =\frac{81\times 912}{7}\\ =10553.14A°\end{array}$

Hence, the required answer is $10553.14A°$