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A nucleus at rest undergoes a decay emitting an α particle 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 / c 2 )

A nucleus at rest undergoes a decay emitting an α  particle 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) 

Grade:upto college level

1 Answers

Navjyot Kalra
askIITians Faculty 654 Points
9 years ago
Hello Student,
Please find the answer to your question
Let the reaction be
Here, my = 223.61 amu and mα = 4.002 amu
We know that
λ = h / mv ⇒ m2 v2 = h2 / λ2 = p2
⇒ But E.K. = p2 / 2m . Therefore K.E. = h2 / 2mλ2 …(i)
Applying eq. (i) for Y and α, we get
K.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 J
Similarly (E.K.)y = 0.0178 x 10-12 J
Total energy = 10-12
We know that E = ∆mc2
∴ ∆m = E /c2 = 10-12 / ( 3 x 108)2 kg
  1. x 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 amu
The mass of the parent nucleus X will be
Mx = my + mα + ∆m
= 223. 61 + 4.002 + 0.00665 = 227.62 amu
Thanks
Navjot Kalra
askIITians Faculty

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