Four lowest energy levels of H-atom are shown | vedclass

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(D) The number of emission lines when an electron transitions from an excited state $n$ to a lower state is given by the formula $\frac{n(n-1)}{2}$,wh

Vedclass · Accepted Answer

$6$

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(D) The number of emission lines when an electron transitions from an excited state $n$ to a lower state is given by the formula $\frac{n(n-1)}{2}$,where $n$ is the principal quantum number of the highest energy level involved.For transitions from $n = 4$ to lower energy levels,the number of possible emission lines is calculated as:$	ext{Number of lines} = \frac{4(4-1)}{2} = \frac{4 	imes 3}{2} = 6$.The possible transitions are:$4 	o 3, 4 	o 2, 4 	o 1, 3 	o 2, 3 	o 1, 2 	o 1$.Thus,there are $6$ possible emission lines.

Four lowest energy levels of $H$-atom are shown in the figure. The number of emission lines could be obtained by transitions from these levels. The total number of possible emission lines is:

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