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(C) The number of spectral lines emitted for transitions from level $n$ to lower levels is given by the formula $N = \frac{n(n-1)}{2}$.Given $N = 10$,

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$\frac{2h}{\pi}$

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(C) The number of spectral lines emitted for transitions from level $n$ to lower levels is given by the formula $N = \frac{n(n-1)}{2}$.Given $N = 10$,we have $\frac{n(n-1)}{2} = 10$,which implies $n^2 - n - 20 = 0$.Solving the quadratic equation,$(n-5)(n+4) = 0$,we get $n = 5$ (since $n$ must be positive).The angular momentum of an electron in the $n$-th orbit is given by $L = \frac{nh}{2\pi}$.The change in angular momentum $\Delta L$ for a transition from $n_2$ to $n_1$ is $\Delta L = \frac{(n_2 - n_1)h}{2\pi}$.To maximize $\Delta L$,we choose the transition from the highest level $n_2 = 5$ to the lowest level $n_1 = 1$.Thus,$\Delta L_{max} = \frac{(5 - 1)h}{2\pi} = \frac{4h}{2\pi} = \frac{2h}{\pi}$.

$A$ sample of hydrogen-like atoms produces an emission spectrum consisting of $10$ wavelengths arising from all possible transitions. During this process,the maximum angular momentum change for an electron transitioning from a higher energy level to a lower energy level is:

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