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

Option: 1

D


Option: 2

B


Option: 3

C


Option: 4

A


Answers (1)

best_answer

\underset{\substack{\downarrow \\(\mathrm{nm})}}{\lambda}=\frac{h c}{\Delta E}=\frac{1241}{\Delta E(\mathrm{ev})}=\frac{1241}{10}=124.1

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chirag

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