Question

Find the wavelength of the first line of spectral series of $H{e}^{+}$ ion whose interval between extreme lines is $\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2}=27419.25{\text{cm}}^{-1}$
(Take Rydberg's constant $R=109677{\text{cm}}^{-1},\frac{1}{109677}=9.11\times {10}^{-6}cm$)

• A. $4685\stackrel{\circ }{A}$
• B. $46.85\stackrel{\circ }{A}$
• C. $7896\stackrel{\circ }{A}$
• D. $78.96\stackrel{\circ }{A}$

Answer 1

The correct option is A $4685\stackrel{\circ }{A}$

Determination of the series:
Extreme lines means first line and last line.
First line corresponds to transition from $n_1+1$ to $n_1$
The last line corresponds to transition from infinity to $n_1$
$\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2}=R{Z}^2[\frac{1}{n_1^2}-\frac{1}{{\infty }^2}]-R{Z}^2[\frac{1}{n_1^2}-\frac{1}{(n_1+1{)}^2}]=\frac{R{Z}^2}{(n_1+1{)}^2}$
$27419.25=\frac{109677\times 2^2}{(n_1+1{)}^2}$
$(n_1+1)=4$
$n_1=3$

Hence series corresponds to Paschen series of $H{e}^{+}$ ion.

Determination of wavelength of the first line:

$\frac{1}{\lambda }=109677\times 2^2\times [\frac{1}{3^2}-\frac{1}{4^2}]$
$\lambda =46.8514\times {10}^{-8}\text{m}=4685.14\stackrel{\circ }{A}$