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(D) The total number of spectral lines emitted when an electron transitions from $n_2$ to $n_1$ is given by the formula: $N = \frac{(n_2 - n_1)(n_2 -

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(D) The total number of spectral lines emitted when an electron transitions from $n_2$ to $n_1$ is given by the formula: $N = \frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2}$.Given $n_2 = 6$ and $n_1 = 2$,the total number of lines is $N = \frac{(6 - 2)(6 - 2 + 1)}{2} = \frac{4 	imes 5}{2} = 10$.Since the transition ends at $n_1 = 2$,all these lines belong to the Balmer series.The question asks for the number of spectral lines excluding the Balmer series.Therefore,the number of lines excluding the Balmer series is $10 - 10 = 0$.

When an electron transitions from the $6^{th}$ orbit to the $2^{nd}$ orbit in a $H$ atom,what is the total number of spectral lines (excluding the Balmer series) produced?

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