Atomic Physics Question 4
Question 4 - 25 January - Shift 2
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 $(h=6.62 \times 10^{-34} Js)$
(1) B
(2) A
(3) $C$
(4) D
Show Answer
Answer: (4)
Solution:
$\lambda=\frac{hc}{\Delta E}$
$\Delta E_A^{a}=2.2 eV$
$\Delta E_B=5.2 eV$
$\Delta E_C=3 eV$
$\Delta E_D=10 eV$
$\lambda_A=\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{2.2 \times 1.6 \times 10^{-19}}$
$=\frac{12.41 \times 10^{-7}}{2.2} m$
$=\frac{1241}{2.2} nm=564 nm$
$\lambda_B=\frac{1241}{5.2} nm=238.65 nm$
$\lambda_C=\frac{1241}{3} nm=413.66 nm$
$\lambda_D=\frac{1241}{10}=124.1 nm$
