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(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2}\,eV$.For $n=1$,$E_1 = -13.6\,eV$.For $n=2$

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$3$

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(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2}\,eV$.For $n=1$,$E_1 = -13.6\,eV$.For $n=2$,$E_2 = -3.4\,eV$.For $n=3$,$E_3 = -1.51\,eV$.For $n=4$,$E_4 = -0.85\,eV$.The energy required to excite an electron from the ground state $(n=1)$ to the $n^{th}$ state is $\Delta E = E_n - E_1$.For $n=2$,$\Delta E = -3.4 - (-13.6) = 10.2\,eV$.For $n=3$,$\Delta E = -1.51 - (-13.6) = 12.09\,eV$.For $n=4$,$\Delta E = -0.85 - (-13.6) = 12.75\,eV$.Since the incident electron beam has an energy of $12.5\,eV$,it can excite hydrogen atoms up to the $n=3$ state,but not to the $n=4$ state.When the electrons de-excite from $n=3$ to $n=1$,the possible transitions are $3 	o 2$,$2 	o 1$,and $3 	o 1$.The number of spectral lines is given by the formula $N = \frac{n(n-1)}{2}$.For $n=3$,$N = \frac{3(3-1)}{2} = 3$.

$A$ $12.5\,eV$ electron beam is used to bombard gaseous hydrogen at room temperature. The number of spectral lines emitted will be:

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