A hydrogen atom (ionisation potential 13.6 e | vedclass

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(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 first excited state,$n_1 = 2

Vedclass · Accepted Answer

$2.55$

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(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 first excited state,$n_1 = 2$.For the third excited state,$n_2 = 4$.The energy of the emitted photon is given by the difference in energies: $\Delta E = E_{n_2} - E_{n_1} = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) \ eV$.Substituting the values: $\Delta E = 13.6 \left( \frac{1}{2^2} - \frac{1}{4^2} ight) = 13.6 \left( \frac{1}{4} - \frac{1}{16} ight) = 13.6 \left( \frac{4-1}{16} ight) = 13.6 	imes \frac{3}{16} = 2.55 \ eV$.

$A$ hydrogen atom (ionisation potential $13.6 \ eV$) makes a transition from the third excited state to the first excited state. The energy of the photon emitted in the process is.......$eV$.

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