Question

The wave number of the first line in the Balmer series of hydrogen atom is $15200c{m}^{-1}$. What is the wave number of the first line in the Balmer series of $B{e}^{3+}$?

• A. $2.432\times {10}^{5}c{m}^{-1}$
• B. $15200c{m}^{-1}$
• C. $4\times 115200c{m}^{-1}$
• D. $2\times 15200c{m}^{-1}$

Answer 1

The correct option is A $2.432\times {10}^{5}c{m}^{-1}$

$\frac{1}{\lambda }=\overline{\lambda }\propto z^2$

$\therefore \frac{\overline{{\lambda }_H}}{\overline{{\lambda }_{Be}}}=\frac{{\left(1\right)}^2}{{\left(4\right)}^2}=\frac{1}{16}$

$\therefore \overline{{\lambda }_{Be}}=16\times \overline{{\lambda }_H}=16\times 15200=2.432\times {10}^5{cm}^{-1}$