Atoms & Nuclei

JEE Main level, single correct, medium

121 of 239

Question 121 of 239

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 124.1nm124.1nm?
Given (h=6.62×10–34Jsh=6.62\times {10}^{–34}Js)

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Answer: (D) D

The change in energy is related to wavelength of photon emitted as:
∆E=hcλ=6.62×10−34Js×3×108ms−11.6×10−19J×124.1nm=1242eVnm124.1nm=10eV∆E=\dfrac{hc}{\lambda } =\dfrac{6.62\times {10}^{-34}Js\times 3\times {10}^{8}m{s}^{-1}}{1.6\times {10}^{-19}J\times 124.1nm} =\dfrac{1242eVnm}{124.1nm} =10eV
Hence, option D is correct.

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