Ionisation potential of hydrogen atom is 13.6 | vedclass

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(B) The energy of the ground state of a hydrogen atom is $E_1 = -13.6 eV$.When energy $\Delta E = 12.1 eV$ is supplied,the electron is excited to a hi

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$3$

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(B) The energy of the ground state of a hydrogen atom is $E_1 = -13.6 eV$.When energy $\Delta E = 12.1 eV$ is supplied,the electron is excited to a higher energy level $n$ such that $E_n = E_1 + \Delta E$.$E_n = -13.6 eV + 12.1 eV = -1.5 eV$.Using the formula $E_n = -\frac{13.6}{n^2} eV$,we have $-\frac{13.6}{n^2} = -1.5$.$n^2 = \frac{13.6}{1.5} \approx 9.07$,which implies $n = 3$.The electron is excited to the second excited state $(n = 3)$.The number of spectral lines emitted when an electron transitions from state $n$ to the ground state is given by $N = \frac{n(n-1)}{2}$.Substituting $n = 3$,we get $N = \frac{3(3-1)}{2} = \frac{3 	imes 2}{2} = 3$ lines.

Ionisation potential of hydrogen atom is $13.6 eV$. When hydrogen atoms in ground state are excited by a supply of $12.1 eV$,then the number of spectral lines emitted by hydrogen atoms according to Bohr's theory is

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