An atom emits energy equal to 4 times 10^{-12 | vedclass

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(B) The wavelength $\lambda$ corresponding to energy $\Delta E = 4 	imes 10^{-12} \, erg$ is calculated using the formula $\lambda = \frac{hc}{\Delta

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Visible region

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(B) The wavelength $\lambda$ corresponding to energy $\Delta E = 4 	imes 10^{-12} \, erg$ is calculated using the formula $\lambda = \frac{hc}{\Delta E}$.Substituting the values: $\lambda = \frac{6.626 	imes 10^{-27} \, erg \cdot s 	imes 3 	imes 10^{10} \, cm/s}{4 	imes 10^{-12} \, erg}$.$\lambda = 4.9695 	imes 10^{-5} \, cm = 4969.5 \, \mathring{A}$.The visible region of the electromagnetic spectrum extends from approximately $3800 \, \mathring{A}$ to $7600 \, \mathring{A}$.Since $4969.5 \, \mathring{A}$ falls within this range,the radiation belongs to the visible region.

An atom emits energy equal to $4 \times 10^{-12} \, erg$. To which part of the electromagnetic spectrum does it belong?

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