Question

Wave number of spectral line for a given transition is $xc{m}^{-1}$ for $H{e}^{+}$, then its value for $B{e}^{3+}$ (isoelectronic of $H{e}^{+}$) for same transition is:

• A. $\frac{x}{4}c{m}^{-1}$
• B. $xc{m}^{-1}$
• C. $4xc{m}^{-1}$
• D. $16xc{m}^{-1}$

Answer 1

Answer: (C)

$4xc{m}^{-1}$

$\overline{v}$ (wave number) $=\overline{R_H}{Z}^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

$\overline{v_1}(H{e}^{+},Z=2)=\overline{R_H}(2{)}^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

$\overline{v_1}(B{e}^{+},Z=4)=\overline{R_H}(4{)}^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

$\therefore \frac{v_2}{v_1}=\frac{(4{)}^2}{(2{)}^2}$

$=\frac{16}{4}=4$

$\therefore \overline{v_2}=4\overline{v_1}$

$\overline{v_2}=4xc{m}^{-1}$

Hence, the correct option is $C$