The energy levels of an atom is shown is figure. Which one of these transitions will result in the emission of a photon of wavelength \(124.1\,nm\) ?Given \(\left( h =6.62 \times 10^{-34}\,Js \right)\)

\(\Delta E _{ A }=2.2\,eV\)

\(\Delta E _{ B }=5.2\,eV\)

\(\Delta E _{ C }=3\,eV\)

\(\Delta E _{ D }=10\,eV\)

\(\lambda_{ A }=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{2.2 \times 1.6 \times 10^{-19}}\)

\(=\frac{12.41 \times 10^{-7}}{2.2}\,m\)

\(=\frac{1241}{2.2}\,nm =564\,nm\)

\(\lambda_{ B }=\frac{1241}{5.2}\,nm =238.65\,nm\)

\(\lambda_{ C }=\frac{1241}{3}\,nm =413.66\,nm\)

\(\lambda_{ D }=\frac{1241}{10}=124.1\,nm\)