Question 154: Atoms & Nuclei

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 (h = 6.62 $	imes$ 10$^{-34}$ Js)
![](https://app-content.cdn.examgoal.net/fly/@width/image/1ldufdovm/1b1015b5-4a1f-485b-bc89-bf0294cb67b6/170a4120-a700-11ed-9824-935677f1e097/file-1ldufdovn.png?format=png)

Accepted Answer

$
\begin{aligned}
& \lambda=\frac{\mathrm{hc}}{\Delta \mathrm{E}} \\
& \Delta \mathrm{E}_{\mathrm{A}}=2.2 \mathrm{eV} \\
& \Delta \mathrm{E}_{\mathrm{B}}=5.2 \mathrm{eV} \\
& \Delta \mathrm{E}_{\mathrm{C}}=3 \mathrm{eV} \\
& \Delta \mathrm{E}_{\mathrm{D}}=10 \mathrm{eV} \\
& \lambda_{\mathrm{A}}=\frac{6.62 	imes 10^{-34} 	imes 3 	imes 10^8}{2.2 	imes 1.6 	imes 10^{-19}} \\
& =\frac{12.41 	imes 10^{-7}}{2.2} \mathrm{~m} \\
& =\frac{1241}{2.2} \mathrm{~nm}=564 \mathrm{~nm} \\
& \lambda_{\mathrm{B}}=\frac{1241}{5.2} \mathrm{~nm}=238.65 \mathrm{~nm} \\
& \lambda_{\mathrm{C}}=\frac{1241}{3} \mathrm{~nm}=413.66 \mathrm{~nm} \\
& \lambda_{\mathrm{D}}=\frac{1241}{10}=124.1 \mathrm{~nm}
\end{aligned}
$

Answer

C

Answer

B

Answer

A

Answer

D

Atoms & Nuclei

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 (h = 6.62 \times 10^{-34} Js)

Choose the correct answer: