A nucleus at rest undergoes a decay emitting an α

Question

A nucleus at rest undergoes a decay emitting an αparticle of de-Broglie wavelength λ = 5.76 x 10-15 m. If the mass of the daughter nucleus is 223. 610 amu and that of the α particles is 4.002 amu, determine the total kinetic energy in the final state. Hence, obtain the mass of the parent nucleus in amu. ( 1 amu = 931. 470 MeV / c2)

Navjyot Kalra · Accepted Answer

Hello Student,Please find the answer to your questionLet the reaction beHere, my = 223.61 amu and mα = 4.002 amuWe know that&lambda; = h / mv &rArr; m2 v2 = h2 / &lambda;2 = p2&rArr; But E.K. = p2 / 2m . Therefore K.E. = h2 / 2m&lambda;2&hellip;(i)Applying eq. (i) for Y and α, we getK.E.α = (6.6 x 10-34)2 / 2 x 4.002 x 1.67 x 10-27 x 5.76 x 10-15 x 5.76 x 10-15= 0.0982243 x 10-11 = 0.982 x 1012 JSimilarly (E.K.)y = 0.0178 x 10-12 JTotal energy = 10-12We know that E = ∆mc2&there4; ∆m = E /c2 = 10-12 / ( 3 x 108)2 kgx 10-27 kg = 1 amu∵ 10-12 / (3 x 108)2 kg = 10-12 amu / 1.67 x 10-27 x ( 3 x 108)2= 10-12 amu / 1. 67 x 9 x 10-27 x 1016 = 0.00665 amuThe mass of the parent nucleus X will beMx = my + mα + ∆m= 223. 61 + 4.002 + 0.00665 = 227.62 amuThanksNavjot KalraaskIITians Faculty

Modern Physics> A nucleus at rest undergoes a decay emitt...

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Navjyot Kalra

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