Question

Find the wavelength of the first line of $H{e}^{+}$ ion spectral series whose interval between extreme line is $[\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2}=2.7451\times {10}^{4}c{m}^{-1}]$.

• A. $6089\mathring{A}$
• B. $4298\mathring{A}$
• C. $4689\mathring{A}$
• D. $10101\mathring{A}$

Answer 1

The correct option is C $4689\mathring{A}$

The transition is $n=2\to n=1$

The expression for the wavelength is $\frac{1}{\lambda }=R{Z}^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

Substitute values in the above expression.
$\frac{1}{{\lambda }_1}=109678c{m}^{-1}\times (2{)}^2\times [\frac{1}{1^2}-\frac{1}{{\infty }^2}]$

$\frac{1}{{\lambda }_2}=109678c{m}^{-1}\times (2{)}^2\times [\frac{1}{1^2}-\frac{1}{2^2}]$

$\frac{1}{\lambda }=[\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2}=2.7451\times {10}^{4}c{m}^{-1}]$

Hence, the wavelength is $\lambda =4689\stackrel{o}{A}$