Question

An $e^{-}$ of $H{e}^{+}$ makes a transition and emits $6^{th}$ line of Balmer series. Similar wavelength of radiation is absorbed by hydrogen like specie to give $9^{th}$ line of paschen series in its spectrum. The value of Z of the hydrogen like specie is :

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (C)

3

Balmer $6^{th}$ line (8---2)
Paschen $9^{th}$ line (12---3)
For $H{e}^{+}\frac{1}{\lambda }=R{.2}^2(\frac{1}{2^2}-\frac{1}{8^2})$

For $He$ atomic number is 2.
$\frac{1}{\lambda }=R(\frac{1}{(2/2{)}^2}-\frac{1}{(8/2{)}^2})$ ....(1)
Fo\displaystyle r single electron species having atomic number 't'
$\frac{1}{\lambda }=R{t}^2(\frac{1}{3^2}-\frac{1}{{12}^2})$

Here t corresponds to atomic number Z of the element.
$\frac{1}{\lambda }=R(\frac{1}{(3/t{)}^2}-\frac{1}{(12/t{)}^2})$ ...(2)
Comparing (1) & (2)
$\Rightarrow {\left(\frac{2}{2}\right)}^2={\left(\frac{3}{t}\right)}^2$
$\Rightarrow Z=3$