In a hydrogen atom,which of the following ele | vedclass

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(B) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} 	ext{ eV}$.The energy change for a transi

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From $n = 3$ to $n = 1$

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(B) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} 	ext{ eV}$.The energy change for a transition from $n_i$ to $n_f$ is $\Delta E = E_{n_i} - E_{n_f} = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight) 	ext{ eV}$.Calculating the energy change for each option:$(A)$ $n = 2$ to $n = 1$: $\Delta E = 13.6 \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = 13.6 	imes (1 - 0.25) = 10.2 	ext{ eV}$.$(B)$ $n = 3$ to $n = 1$: $\Delta E = 13.6 \left( \frac{1}{1^2} - \frac{1}{3^2} ight) = 13.6 	imes (1 - 0.111) = 12.09 	ext{ eV}$.$(C)$ $n = 4$ to $n = 2$: $\Delta E = 13.6 \left( \frac{1}{2^2} - \frac{1}{4^2} ight) = 13.6 	imes (0.25 - 0.0625) = 2.55 	ext{ eV}$.$(D)$ $n = 3$ to $n = 2$: $\Delta E = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = 13.6 	imes (0.25 - 0.111) = 1.89 	ext{ eV}$.Comparing these values,the transition from $n = 3$ to $n = 1$ involves the maximum energy change of $12.09 	ext{ eV}$. Thus,option $(B)$ is correct.

In a hydrogen atom,which of the following electronic transitions would involve the maximum energy change?

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