HomeLearn Fast select Course
Show Unit/ChapterChange UnitChange SubjectLast Menu
NEET-XII-Physics

46: The Nucleus

with Solutions - page 5
Qstn# iv-18 Prvs-QstnNext-Qstn
  • #18
    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.
    Ans : 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 `` {\beta }^{+}`` decay will be
    Qvalue= [ m(12N) `` -`` (m(12C*) + 2me)]c2
    = [12.018613 `` \times ``931 MeV `` -`` (12`` \times ``931 + 4.43) MeV `` -`` (2`` \times ``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.
    Page No 442: