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 $124.1nm$ ?
Given $\left(h=6.62\times 10^{-34}Js\right)$

• A. D
• B. B
• C. A
• D. $C$

Answer 1

Answer: (A)

$\lambda =\frac{hc}{\Delta E}$
$\Delta E_A=2.2eV$
$\Delta E_B=5.2eV$
$\Delta E_C=3eV$
$\Delta E_D=10eV$
${\lambda }_A=\frac{6.62\times 10^{-34}\times 3\times 10^8}{2.2\times 1.6\times 10^{-19}}$
$=\frac{12.41\times 10^{-7}}{2.2}m$
$=\frac{1241}{2.2}nm=564nm$
${\lambda }_B=\frac{1241}{5.2}nm=238.65nm$
${\lambda }_C=\frac{1241}{3}nm=413.66nm$
${\lambda }_D=\frac{1241}{10}=124.1nm$