The ionization potential of a hydrogen atom i | vedclass

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(C) The ionization potential of a hydrogen atom is $13.6 \ eV$. The energy of the $n^{th}$ orbit is given by $E_n = -\frac{13.6}{n^2} \ eV$.When the h

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$3$

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(C) The ionization potential of a hydrogen atom is $13.6 \ eV$. The energy of the $n^{th}$ orbit is given by $E_n = -\frac{13.6}{n^2} \ eV$.When the hydrogen atom in the ground state $(n=1)$ absorbs a photon of energy $12.1 \ eV$,it is excited to a higher energy level $n$. The energy difference is given by:$E = E_n - E_1$$12.1 = -\frac{13.6}{n^2} - (-13.6)$$12.1 = 13.6 - \frac{13.6}{n^2}$$\frac{13.6}{n^2} = 13.6 - 12.1 = 1.5$$n^2 = \frac{13.6}{1.5} \approx 9.06 \approx 9$$n = 3$The number of spectral lines emitted when the electron transitions from the $n^{th}$ state back to the ground state is given by the formula $\frac{n(n-1)}{2}$.For $n=3$,the number of spectral lines = $\frac{3(3-1)}{2} = \frac{3 	imes 2}{2} = 3$.

The ionization potential of a hydrogen atom is $13.6 \ eV$. Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy $12.1 \ eV$. According to Bohr's theory,the number of spectral lines emitted by the hydrogen atoms will be:

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