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A doubly ionized lithium atom is hydrogen-like with atomic number $3$ is excited to third energy state. How many spectral lines are observed in the emission spectrum of the above excited system?

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$N = 3$

$N = 5$

$N = 6$

$N = 7$

answer is A.

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Detailed Solution

Given, $Z = 3$

Since, $E_{n} \propto \frac{Z^{2}}{n^{2}}$

From $n^{th}$ orbit, total number of emission lines $(N)$ can be $N = \frac{n (n - 1)}{2} = \frac{3 (3 - 1)}{2} = 3 .$

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