When a hydrogen atom emits a photon of energy | vedclass

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(C) The energy of an electron in the $n^{	ext{th}}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \,eV$.When the atom emits a photon

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$\frac{h}{\pi}$

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(C) The energy of an electron in the $n^{	ext{th}}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \,eV$.When the atom emits a photon of energy $12.09 \,eV$,the electron jumps from a higher energy level $n_2$ to a lower energy level $n_1$.The energy difference is $\Delta E = E_{n_2} - E_{n_1} = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) = 12.09 \,eV$.Dividing by $13.6$,we get $\frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{12.09}{13.6} \approx 0.889$.For $n_1 = 1$,we have $1 - \frac{1}{n_2^2} = 0.889$,which gives $\frac{1}{n_2^2} = 0.111$,so $n_2^2 = 9$,implying $n_2 = 3$.Thus,the electron jumps from the $3^{	ext{rd}}$ orbit to the $1^{	ext{st}}$ orbit.The orbital angular momentum is given by $L = \frac{nh}{2\pi}$.The change in angular momentum is $\Delta L = L_2 - L_1 = \frac{n_2 h}{2\pi} - \frac{n_1 h}{2\pi} = \frac{(3-1)h}{2\pi} = \frac{2h}{2\pi} = \frac{h}{\pi}$.

When a hydrogen atom emits a photon of energy $12.09 \,eV$,its orbital angular momentum changes by (where $h$ is Planck's constant)

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