When a hydrogen atom (ionization energy 13.6 | vedclass

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Question

(B) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \ eV$.For the third excited state,$n_2 = 4

Vedclass · Accepted Answer

$2.55$

Vedclass · Answer

(B) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \ eV$.For the third excited state,$n_2 = 4$.For the first excited state,$n_1 = 2$.The energy of the emitted photon is $\Delta E = E_{n_2} - E_{n_1} = -13.6 \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} ight)$.Substituting the values: $\Delta E = 13.6 \left( \frac{1}{2^2} - \frac{1}{4^2} ight) = 13.6 \left( \frac{1}{4} - \frac{1}{16} ight)$.$\Delta E = 13.6 \left( \frac{4-1}{16} ight) = 13.6 	imes \frac{3}{16} = 2.55 \ eV$.

When a hydrogen atom (ionization energy $13.6 \ eV$) jumps from the third excited state to the first excited state,the energy of the emitted photon in this process is .............. $eV$.

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