When the potential of the second excited stat | vedclass

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(C) The energy of the photon emitted is $E = 108.8 \ eV = 108.8 	imes 1.6 	imes 10^{-19} \ J$.The momentum of the photon is given by $p = \frac{E}{c

Vedclass · Accepted Answer

$1.44 	imes 10^{-25}$

Vedclass · Answer

(C) The energy of the photon emitted is $E = 108.8 \ eV = 108.8 	imes 1.6 	imes 10^{-19} \ J$.The momentum of the photon is given by $p = \frac{E}{c} = \frac{108.8 	imes 1.6 	imes 10^{-19}}{3 	imes 10^8} \approx 5.8 	imes 10^{-26} \ kg \cdot m/s$.By the law of conservation of momentum,the recoil momentum of the atom is equal to the momentum of the photon,$p_{atom} = p_{photon} = 5.8 	imes 10^{-26} \ kg \cdot m/s$.Assuming the atom is a Lithium atom (as $108.8 \ eV$ corresponds to the energy transition in $Li^{++}$),the mass $M \approx 7 	imes 1.67 	imes 10^{-27} \ kg$.The recoil energy $E_r$ is given by $E_r = \frac{p^2}{2M} = \frac{(5.8 	imes 10^{-26})^2}{2 	imes 7 	imes 1.67 	imes 10^{-27}} \approx 1.44 	imes 10^{-25} \ J$.

When the potential of the second excited state of an atom is $108.8 \ V$,the recoil energy of the atom is ...... Joules.

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