The ionization potential of a hydrogen atom i | vedclass

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(C) The energy of the electron in the ground state $(n=1)$ of a hydrogen atom is $E_1 = -13.6 \ eV$.When the atom absorbs a photon of energy $12.1 \ e

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$3$

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(C) The energy of the electron in the ground state $(n=1)$ of a hydrogen atom is $E_1 = -13.6 \ eV$.When the atom absorbs a photon of energy $12.1 \ eV$,the final energy of the electron becomes $E_f = E_1 + 12.1 \ eV = -13.6 \ eV + 12.1 \ eV = -1.51 \ eV$.Using the formula $E_n = -\frac{13.6}{n^2} \ eV$,we find the excited state $n$:$-1.51 = -\frac{13.6}{n^2} \implies n^2 = \frac{13.6}{1.51} \approx 9 \implies n = 3$.The number of spectral lines emitted when an electron transitions from an excited state $n$ to lower energy states is given by the formula $N = \frac{n(n-1)}{2}$.For $n=3$,the number of spectral lines is $N = \frac{3(3-1)}{2} = \frac{3 	imes 2}{2} = 3$.

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 $12.1 \ eV$. The number of spectral lines emitted by the hydrogen atoms according to Bohr's theory will be

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