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(C) The number of dark lines in an absorption spectrum is given by $(n - 1)$,where $n$ is the principal quantum number of the highest excited state re

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$15$

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(C) The number of dark lines in an absorption spectrum is given by $(n - 1)$,where $n$ is the principal quantum number of the highest excited state reached by the atoms.Given that there are $5$ dark lines,we have $n - 1 = 5$,which implies $n = 6$.When these atoms return to the ground state,the number of bright lines in the emission spectrum is given by the formula $\frac{n(n - 1)}{2}$.Substituting $n = 6$ into the formula,we get $\frac{6(6 - 1)}{2} = \frac{6 	imes 5}{2} = 15$.Therefore,the number of bright lines in the emission spectrum is $15$.

In a sample of hydrogen-like atoms,all of which are in the ground state,a photon beam containing photons of various energies is passed. In the absorption spectrum,five dark lines are observed. The number of bright lines in the emission spectrum will be (assume that all transitions take place).

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