Question

The wavenumber of the first line in Balmer series of H - atom is 15,200 ${cm}^{-1}$, the wavenumber of the first line in Balmer series of ${Be}^{3+}$:

• A. $2.43\times {10}^5{cm}^{-1}$
• B. $3.43\times {10}^5{cm}^{-1}$
• C. $4.43\times {10}^5{cm}^{-1}$
• D. $5.43\times {10}^5{cm}^{-1}$

Answer 1

The correct option is A $2.43\times {10}^5{cm}^{-1}$

$Wavenumber\propto \frac{Z^2}{n^2}$
$Be\Rightarrow Z=4$

$\frac{{\nu }_1}{{\nu }_2}=\frac{{Z_1}^2}{{Z_2}^2}$

Wave number of $1^{st}$ line in the Balmer series of $B{e}^{3+}=16\times 15200c{m}^{-1}$

$=2.43\times {10}^{5}c{m}^{-1}$

Hence, option A is correct.