A nucleus of mass 20 u emits a gamma phot | vedclass

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(B) Given: Mass of nucleus $ M = 20 u = 20 	imes 1.6 	imes 10^{-27} kg = 3.2 	imes 10^{-26} kg $.Energy of photon $ E = 6 MeV = 6 	imes 10^6 eV =

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(B) Given: Mass of nucleus $ M = 20 u = 20 	imes 1.6 	imes 10^{-27} kg = 3.2 	imes 10^{-26} kg $.Energy of photon $ E = 6 MeV = 6 	imes 10^6 eV = 6 	imes 10^6 	imes 1.6 	imes 10^{-19} J = 9.6 	imes 10^{-13} J $.By conservation of linear momentum, the momentum of the nucleus $ p_n $ must be equal to the momentum of the photon $ p_p $.$ p_p = \frac{E}{c} = \frac{9.6 	imes 10^{-13} J}{3 	imes 10^8 m/s} = 3.2 	imes 10^{-21} kg \cdot m/s $.Since $ p_n = p_p $, the kinetic energy of the nucleus $ K_n $ is given by $ K_n = \frac{p_n^2}{2M} $.$ K_n = \frac{(3.2 	imes 10^{-21})^2}{2 	imes 3.2 	imes 10^{-26}} = \frac{10.24 	imes 10^{-42}}{6.4 	imes 10^{-26}} = 1.6 	imes 10^{-16} J $.Converting to $ eV $: $ K_n = \frac{1.6 	imes 10^{-16} J}{1.6 	imes 10^{-19} J/eV} = 10^3 eV = 1 keV $.Thus, the kinetic energy of the nucleus is $ 1 keV $.

$A$ nucleus of mass $ 20 u $ emits a $ \gamma $ photon of energy $ 6 MeV $. If the emission is assumed to occur when the nucleus is free and at rest, then the nucleus will have a kinetic energy nearest to (take $ 1 u = 1.6 \times 10^{-27} kg $): (in $keV$)

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