Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
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Calculate the Maximum Kinetic Energy of the Beta Particle Emitted in the Following Decay Scheme: - Physics

Question

Calculate the maximum kinetic energy of the beta particle emitted in the following decay scheme:
12N → 12C* + e+ + v
12C* → 12C + γ (4.43MeV).
The atomic mass of 12N is 12.018613 u.

(Use Mass of proton mp = 1.007276 u, Mass of ""_1^1"H" atom = 1.007825 u, Mass of neutron mn = 1.008665 u, Mass of electron = 0.0005486 u ≈ 511 keV/c2,1 u = 931 MeV/c2.)

Solution

Given:-
Atomic mass of 12N, m(12N) = 12.018613 u
12N → 12C* + e+ + v
12C* → 12C + γ (4.43 MeV)

Net reaction is given by

12N → 12C + e+ + v + γ (4.43 MeV)

Qvalue  of the β^+ decay will be

Qvalue = [m(""^12N) - (m(12C*) + 2me)]c2

= [12.018613 xx 931 "MeV" - (12 xx 931 + 4.43) "MeV" - (2 xx 511) "keV"]

= [11189.3287  - 11176.43  - 1.022] MeV

= 11.8767  "MeV" = 11.88  "MeV"

The maximum kinetic energy of beta particle will be 11.88 MeV, assuming that neutrinos have zero energy.

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Solution Calculate the Maximum Kinetic Energy of the Beta Particle Emitted in the Following Decay Scheme: Concept: Radioactivity - Law of Radioactive Decay.
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