The ionisation potential of H-atom is 13.6 , | vedclass

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(C) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Substituting the values: $E = \frac{12400 \, eV \cdot \mathring{A}}{970.6 \

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$6$

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(C) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Substituting the values: $E = \frac{12400 \, eV \cdot \mathring{A}}{970.6 \, \mathring{A}} \approx 12.77 \, eV$.The energy of the $n$-th state in a Hydrogen atom is $E_n = -\frac{13.6}{n^2} \, eV$.The energy required to excite from ground state $(n=1)$ to state $n$ is $\Delta E = 13.6 \left( 1 - \frac{1}{n^2} ight)$.Setting $\Delta E = 12.77 \, eV$,we get $1 - \frac{1}{n^2} = \frac{12.77}{13.6} \approx 0.939$.$\frac{1}{n^2} = 1 - 0.939 = 0.061 \Rightarrow n^2 \approx 16 \Rightarrow n = 4$.The number of emission lines for an electron transitioning from state $n$ to lower states is given by $N = \frac{n(n-1)}{2}$.For $n=4$,$N = \frac{4(4-1)}{2} = \frac{4 	imes 3}{2} = 6$.

The ionisation potential of $H$-atom is $13.6 \, eV$. When it is excited from the ground state by monochromatic radiation of $970.6 \, \mathring{A}$,the number of emission lines will be (according to Bohr's theory):

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