What will be the frequency of the radiation e | vedclass

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(B) The energy of an electron in the $n^{th}$ orbit is given by $E_n = -2.18 	imes 10^{-18} \,J 	imes \frac{1}{n^2}$.For $n = 1$,$E_1 = -2.18 	imes

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$3.08 	imes 10^{15} \,s^{-1}$

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(B) The energy of an electron in the $n^{th}$ orbit is given by $E_n = -2.18 	imes 10^{-18} \,J 	imes \frac{1}{n^2}$.For $n = 1$,$E_1 = -2.18 	imes 10^{-18} \,J$.For $n = 4$,$E_4 = \frac{-2.18 	imes 10^{-18}}{4^2} = -0.13625 	imes 10^{-18} \,J$.The energy difference is $\Delta E = E_4 - E_1 = (-0.13625 	imes 10^{-18}) - (-2.18 	imes 10^{-18}) = 2.04375 	imes 10^{-18} \,J$.Using the relation $\Delta E = h 
u$,the frequency $
u$ is $
u = \frac{\Delta E}{h} = \frac{2.04375 	imes 10^{-18} \,J}{6.625 	imes 10^{-34} \,J \cdot s} \approx 3.08 	imes 10^{15} \,s^{-1}$.

What will be the frequency of the radiation emitted when an electron in a hydrogen atom jumps from $n = 4$ to $n = 1$? $(h = 6.625 \times 10^{-34} \,J \cdot s)$

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