The ratio of the kinetic energies of the elec | vedclass

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(C) The kinetic energy $(K)$ of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $K_n = \frac{13.6 	ext{ eV}}{n^2}$.For the third exc

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$25: 16$

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(C) The kinetic energy $(K)$ of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $K_n = \frac{13.6 	ext{ eV}}{n^2}$.For the third excited state,the principal quantum number is $n_1 = 3 + 1 = 4$.For the fourth excited state,the principal quantum number is $n_2 = 4 + 1 = 5$.The kinetic energy in the third excited state is $K_4 = \frac{13.6}{4^2} = \frac{13.6}{16}$.The kinetic energy in the fourth excited state is $K_5 = \frac{13.6}{5^2} = \frac{13.6}{25}$.The ratio of the kinetic energies is $\frac{K_4}{K_5} = \frac{13.6/16}{13.6/25} = \frac{25}{16}$.Thus,the ratio is $25: 16$.

The ratio of the kinetic energies of the electrons in the third and fourth excited states of the hydrogen atom is

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