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(C) The longest wavelength of the Balmer series for a hydrogen atom corresponds to the transition of an electron from the $n=3$ state to the $n=2$ sta

Vedclass · Accepted Answer

$12.1$

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(C) The longest wavelength of the Balmer series for a hydrogen atom corresponds to the transition of an electron from the $n=3$ state to the $n=2$ state.This implies that when the hydrogen atom was excited by the colliding electron,the electron was promoted from the $n=1$ ground state to the $n=3$ excited state.The energy of the $n$-th quantum state for a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} 	ext{ eV}$.For the ground state $(n=1)$: $E_1 = -\frac{13.6}{1^2} = -13.6 	ext{ eV}$.For the excited state $(n=3)$: $E_3 = -\frac{13.6}{3^2} = -1.51 	ext{ eV}$.The energy required for this excitation is $\Delta E = E_3 - E_1 = -1.51 - (-13.6) = 12.09 	ext{ eV} \approx 12.1 	ext{ eV}$.Since the colliding electron loses all its kinetic energy to excite the atom,the minimum kinetic energy of the electron must be equal to the excitation energy,which is $12.1 	ext{ eV}$.

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