The four lowest energy levels of the H atom a | vedclass

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(D) For an electron transitioning from an excited state $n$ to lower energy levels,the total number of possible emission lines is given by the formula

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(D) For an electron transitioning from an excited state $n$ to lower energy levels,the total number of possible emission lines is given by the formula: $\frac{n(n - 1)}{2}$.Given that the highest energy level shown is $n = 4$,we substitute $n = 4$ into the formula:$	ext{Number of lines} = \frac{4(4 - 1)}{2} = \frac{4 	imes 3}{2} = 6$.Therefore,the total number of emission lines is $6$.

The four lowest energy levels of the $H$ atom are shown in the figure. The number of emission lines will be ...... .

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