Question

Calculate the energy of radiation emitted for the electronic transition from infinity to ground state for $L{i}^{2+}$ ion.
$(\text{Given}:c=3\times {10}^8\text{m/s},R_H=1.09678\times {10}^7{\text{m}}^{-1}$
$h=6.626\times {10}^{-34}{\text{J s}}^{-1})$

• A. $19.66\times {10}^{-18}\text{J}$
• B. $17.66\times {10}^{-18}\text{J}$
• C. $9.83\times {10}^{-18}\text{J}$
• D. $8.83\times {10}^{-18}\text{J}$

Answer 1

The correct option is A $19.66\times {10}^{-18}\text{J}$

From Rydberg equation, wavenumber
$\stackrel{-}{v}=\frac{1}{\lambda }=R_H{Z}^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

when $n_1=1,n_2=\infty ,Z=3$
$\frac{1}{\lambda }=1.09678\times {10}^7\times 3^2[\frac{1}{1^2}-\frac{1}{{\infty }^2}]$

$\Rightarrow \lambda =1.01\times {10}^{-8}\text{m}=101\stackrel{o}{\text{A}}$

Energy associated with this radiation,
$E=\frac{hc}{\lambda }=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{101\times {10}^{-10}}$

$=19.66\times {10}^{-18}\text{J}$