Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
Share
Notifications

View all notifications

Calculate the Maximum Kinetic Energy of the Beta Particle Emitted in the Following Decay Scheme: - Physics

Login
Create free account


      Forgot password?

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.

  Is there an error in this question or solution?

APPEARS IN

Video TutorialsVIEW ALL [1]

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