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(C) The electron in the ground state of the $H$-atom jumps to the $n^{th}$ state after absorbing the radiation.Wavelength of the radiation,$\lambda =

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$6$

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(C) The electron in the ground state of the $H$-atom jumps to the $n^{th}$ state after absorbing the radiation.Wavelength of the radiation,$\lambda = 970 \mathring A = 970 	imes 10^{-10} 	ext{ m}$.Energy gained by the electron,$E' = \frac{hc}{e\lambda} = \frac{1.237 	imes 10^{-6}}{970 	imes 10^{-10}} 	ext{ eV} = 12.75 	ext{ eV}$.Thus,the energy of the $n^{th}$ state,$E_n = -13.6 + 12.75 = -0.85 	ext{ eV}$.Using the formula $E_n = \frac{-13.6}{n^2} 	ext{ eV}$:$-0.85 = \frac{-13.6}{n^2}$$n^2 = \frac{13.6}{0.85} = 16$$n = 4$.The number of emission spectral lines is given by $N = \frac{n(n-1)}{2}$.$N = \frac{4(4-1)}{2} = \frac{4 	imes 3}{2} = 6$ lines.

$A$ hydrogen atom in its ground state is irradiated by light of wavelength $970 \mathring A$. Taking $hc/e = 1.237 \times 10^{-6} \text{ eV m}$ and the ground state energy of the hydrogen atom as $-13.6 \text{ eV}$,the number of lines present in the emission spectrum is

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