When an electron in an excited atom jumps fro | vedclass

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(B) Using Bragg's Law for first-order diffraction $(n=1)$: $n \lambda = 2d \sin 	heta$ $\lambda = 2 	imes (2.64 	imes 10^{-10} \ m) 	imes \sin(7.7

Vedclass · Accepted Answer

$2.788 	imes 10^{-15} \ J$

Vedclass · Answer

(B) Using Bragg's Law for first-order diffraction $(n=1)$: $n \lambda = 2d \sin 	heta$ $\lambda = 2 	imes (2.64 	imes 10^{-10} \ m) 	imes \sin(7.75^{\circ})$ $\lambda = 5.28 	imes 10^{-10} 	imes 0.1349 \approx 7.123 	imes 10^{-11} \ m$ The energy difference $\Delta E$ is given by: $\Delta E = \frac{hc}{\lambda} = \frac{(6.626 	imes 10^{-34} \ J \cdot s) 	imes (3 	imes 10^8 \ m/s)}{7.123 	imes 10^{-11} \ m}$ $\Delta E \approx 2.79 	imes 10^{-15} \ J$ Thus,the closest option is $2.788 	imes 10^{-15} \ J$.

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