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(D) Given equations: $2n_2 + 3n_1 = 18$ $(I)$ $2n_2 - 3n_1 = 6$ $(II)$ Adding equations $(I)$ and $(II)$: $4n_2 = 24 \implies n_2 = 6$ $(III)$ Substit

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$10$

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(D) Given equations: $2n_2 + 3n_1 = 18$ $(I)$ $2n_2 - 3n_1 = 6$ $(II)$ Adding equations $(I)$ and $(II)$: $4n_2 = 24 \implies n_2 = 6$ $(III)$ Substituting the value of $(III)$ into $(II)$: $2(6) - 3n_1 = 6 \implies 12 - 3n_1 = 6 \implies 3n_1 = 6 \implies n_1 = 2$ Thus,the transition occurs between orbits $6 ightarrow 2$. The formula to find the total number of emitted photons is: $	ext{Total photons} = \frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2}$ $= \frac{(6 - 2)(6 - 2 + 1)}{2} = \frac{4 	imes 5}{2} = 10$ Therefore,the correct option is $D$.

In a $He^{+}$ ion,an electronic transition occurs from orbit $n_2$ to $n_1$ such that:
$2n_2 + 3n_1 = 18$
$2n_2 - 3n_1 = 6$
What is the total number of photons emitted when the electron transitions to the $n_1$ orbit?

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In a $He^{+}$ ion,an electronic transition occurs from orbit $n_2$ to $n_1$ such that: $2n_2 + 3n_1 = 18$ $2n_2 - 3n_1 = 6$ What is the total number of photons emitted when the electron transitions to the $n_1$ orbit?