The recoil speed of a hydrogen atom after it | vedclass

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(A) The energy released when an electron transitions from $n = 5$ to $n = 1$ is given by:$\Delta E = E_5 - E_1 = -0.54 \, eV - (-13.6 \, eV) = 13.06 \

Vedclass · Accepted Answer

$4.17$

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(A) The energy released when an electron transitions from $n = 5$ to $n = 1$ is given by:$\Delta E = E_5 - E_1 = -0.54 \, eV - (-13.6 \, eV) = 13.06 \, eV$.By the law of conservation of linear momentum,the initial momentum of the atom is zero,so the final momentum of the atom must be equal and opposite to the momentum of the emitted photon:$P_{	ext{atom}} = P_{	ext{photon}} = \frac{h}{\lambda}$.Since the energy of the photon is $\Delta E = \frac{hc}{\lambda}$,we have $\frac{1}{\lambda} = \frac{\Delta E}{hc}$.Substituting this into the momentum equation,the recoil momentum of the atom is $Mv = \frac{\Delta E}{c}$,where $M$ is the mass of the hydrogen atom $(M \approx 1.67 	imes 10^{-27} \, kg)$ and $c$ is the speed of light $(3 	imes 10^8 \, m/s)$.The recoil speed $v$ is given by:$v = \frac{\Delta E}{Mc} = \frac{13.06 	imes 1.6 	imes 10^{-19} \, J}{1.67 	imes 10^{-27} \, kg 	imes 3 	imes 10^8 \, m/s} \approx 4.17 \, m/s$.

The recoil speed of a hydrogen atom after it emits a photon in going from $n = 5$ state to $n = 1$ state will be ..... $m/s$.

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