The minimum energy required by a hydrogen ato | vedclass

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(D) To emit radiation in the Balmer series,an electron must transition to the $n=2$ energy level from a higher energy level $(n > 2)$.For the minimum

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$12.1$

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(D) To emit radiation in the Balmer series,an electron must transition to the $n=2$ energy level from a higher energy level $(n > 2)$.For the minimum energy,the electron must be excited from the ground state $(n=1)$ to the lowest possible excited state that allows a transition to $n=2$,which is $n=3$.The energy of the $n$-th state of a hydrogen atom is given by $E_n = -\frac{13.6 \ eV}{n^2}$.The energy required to excite the electron from $n=1$ to $n=3$ is $\Delta E = E_3 - E_1$.$E_1 = -13.6 \ eV$.$E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \approx -1.51 \ eV$.$\Delta E = -1.51 - (-13.6) = 12.09 \ eV \approx 12.1 \ eV$.

The minimum energy required by a hydrogen atom in the ground state to emit radiation in the Balmer series is nearly: (in $eV$)

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