Question

The energy levels of an atom is shown in figure.

Which one of these transitions will result in the emission of a photon of wavelength $124.1\, nm$ ?
Given $\left( h =6.62 \times 10^{-34} Js \right)$

• A. D
• B. B
• C. A
• D. $C$

Answer 1

Answer: (A)

$ \lambda=\frac{ hc }{\Delta E } $
$ \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$