JEE MAIN - Physics (2023 - 25th January Evening Shift - No. 19)
The energy levels of an atom is shown in figure.
_25th_January_Evening_Shift_en_19_1.png)
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)
C
B
A
D
Explanation
$$
\begin{aligned}
& \lambda=\frac{\mathrm{hc}}{\Delta \mathrm{E}} \\\\
& \Delta \mathrm{E}_{\mathrm{A}}=2.2 \mathrm{eV} \\\\
& \Delta \mathrm{E}_{\mathrm{B}}=5.2 \mathrm{eV} \\\\
& \Delta \mathrm{E}_{\mathrm{C}}=3 \mathrm{eV} \\\\
& \Delta \mathrm{E}_{\mathrm{D}}=10 \mathrm{eV} \\\\
& \lambda_{\mathrm{A}}=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{2.2 \times 1.6 \times 10^{-19}} \\\\
& =\frac{12.41 \times 10^{-7}}{2.2} \mathrm{~m} \\\\
& =\frac{1241}{2.2} \mathrm{~nm}=564 \mathrm{~nm} \\\\
& \lambda_{\mathrm{B}}=\frac{1241}{5.2} \mathrm{~nm}=238.65 \mathrm{~nm} \\\\
& \lambda_{\mathrm{C}}=\frac{1241}{3} \mathrm{~nm}=413.66 \mathrm{~nm} \\\\
& \lambda_{\mathrm{D}}=\frac{1241}{10}=124.1 \mathrm{~nm}
\end{aligned}
$$
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