The electron in a hydrogen atom in a sample i | vedclass

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(B) The $n^{th}$ excited state corresponds to the principal quantum number $N = n + 1$.When an electron transitions from an excited state $N$ to lower

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$1 + 2 + 3 + \dots + n$

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(B) The $n^{th}$ excited state corresponds to the principal quantum number $N = n + 1$.When an electron transitions from an excited state $N$ to lower energy levels,the total number of spectral lines emitted is given by the formula $\frac{N(N - 1)}{2}$.Substituting $N = n + 1$ into the formula,we get:Number of lines $= \frac{(n + 1)((n + 1) - 1)}{2} = \frac{(n + 1)n}{2}$.This expression is equivalent to the sum of the first $n$ natural numbers: $1 + 2 + 3 + \dots + n$.

The electron in a hydrogen atom in a sample is in the $n^{th}$ excited state. The number of different spectral lines obtained in its emission spectrum will be:

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