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(C) According to Einstein's mass-energy equivalence, the energy difference $\Delta E$ between the excited state and the ground state corresponds to a

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$10^{40}$

Vedclass · Answer

(C) According to Einstein's mass-energy equivalence, the energy difference $\Delta E$ between the excited state and the ground state corresponds to a mass difference $\Delta m$ given by $\Delta E = \Delta m c^2$.Given the excitation frequency $
u = 9.2 	imes 10^9 \, Hz$, the energy difference is $\Delta E = h
u$.Therefore, the mass difference for one pair of atoms is $\Delta m = \frac{h
u}{c^2}$.Substituting the given values:$\Delta m = \frac{6.63 	imes 10^{-34} 	imes 9.2 	imes 10^9}{(3 	imes 10^8)^2} \, kg$$\Delta m = \frac{6.63 	imes 9.2 	imes 10^{-25}}{9 	imes 10^{16}} \, kg$$\Delta m \approx 6.78 	imes 10^{-41} \, kg$.Since this mass difference is obtained using $2$ atoms, the number of atoms $N$ required to obtain a total mass of $1 \, kg$ is:$N = \frac{2 	imes 1 \, kg}{\Delta m} = \frac{2}{6.78 	imes 10^{-41}} \approx 0.295 	imes 10^{41} \approx 3 	imes 10^{40}$.Thus, the order of magnitude is $10^{40}$.

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