Question

Calculate the energy of radiation emitted for the electronic transition from infinity $\left(\infty \right)$ to ground state of an H-atom.
Rydberg constant = $1.09678\times {10}^7{m}^{-1}$

• A. $2.18\times {10}^{-20}$ J
• B. $1.81\times {10}^{-20}J$
• C. $2.18\times {10}^{-18}J$
• D. $1.81\times {10}^{-18}J$

Answer 1

The correct option is C $2.18\times {10}^{-18}J$

Using the formula,
$\frac{1}{\lambda }=R_H[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
Where $\lambda$ = wavelength of the radiation
$n_1=1and{n}_2=\infty$
$\frac{1}{\lambda }=1.09678\times {10}^7{m}^{-1}[\frac{1}{1^2}-\frac{1}{{\infty }^2}]=1.09678\times {10}^7{m}^{-1}\lambda =9.11\times {10}^{-8}mE=\frac{hc}{\lambda }=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{9.11\times {10}^{-8}}=2.18\times {10}^{-18}J$