Question

If the shortest wavelength of $H$-atom in Lyman series is $x$, then longest wavelength in Balmer series of $H{e}^{+}$is :-

• A. $\frac{9x}{5}$
• B. $\frac{36x}{5}$
• C. $\frac{x}{4}$
• D. $\frac{5x}{9}$

Answer 1

Answer: (A)

$\frac{1}{\lambda }=R_H\times Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
$\frac{1}{{\lambda }_1}=R_H\times (1)\left(\frac{1}{(1{)}^2}-\frac{1}{(\infty {)}^2}\right)$
${\lambda }_1=\frac{1}{R_H}=X$
$\Rightarrow R_H=\frac{1}{X}$
$\frac{1}{{\lambda }_2}=R_H\times (2{)}^2\left(\frac{1}{(2{)}^2}-\frac{1}{(3{)}^2}\right)$
$\frac{1}{{\lambda }_2}=R_H\times 4\left(\frac{1}{4}-\frac{1}{9}\right)$
$\frac{1}{{\lambda }_2}=R_H\times 4\left(\frac{9-4}{4\times 9}\right)$
${\lambda }_2=\frac{9}{5}{R}_H$
$\Rightarrow {\lambda }_2=\frac{9}{5}X$