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(A) Given: Wavelength of the photon $\lambda = 0.1 \ \mathring{A} = 10^{-11} \ m$. Using the law of conservation of momentum, the momentum of the atom

Vedclass · Accepted Answer

$2.04$

Vedclass · Answer

(A) Given: Wavelength of the photon $\lambda = 0.1 \ \mathring{A} = 10^{-11} \ m$. Using the law of conservation of momentum, the momentum of the atom must be equal to the momentum of the emitted photon: $p_{atom} = p_{photon} = \frac{h}{\lambda}$. The recoil kinetic energy $(K.E.)$ of the atom is given by $K.E. = \frac{p_{atom}^2}{2m} = \frac{h^2}{2\lambda^2 m}$, where $m$ is the mass of the helium atom. The mass of a helium atom is $m = \frac{4 	imes 10^{-3} \ kg}{6.023 	imes 10^{23}} \approx 6.64 	imes 10^{-27} \ kg$. Substituting the values: $K.E. = \frac{(6.626 	imes 10^{-34})^2}{2 	imes (10^{-11})^2 	imes 6.64 	imes 10^{-27}} \ J$. $K.E. \approx \frac{43.9 	imes 10^{-68}}{13.28 	imes 10^{-49}} \ J \approx 3.3 	imes 10^{-19} \ J$. Converting to $eV$: $K.E. = \frac{3.3 	imes 10^{-19}}{1.6 	imes 10^{-19}} \ eV \approx 2.06 \ eV$. Rounding to the nearest provided option, the correct answer is $2.04 \ eV$.

The wavelength of a photon emitted by a stationary helium atom is $0.1 \ \mathring{A}$. What is the recoil energy of the atom in $eV$?

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