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प्रश्न
23Ne decays to 23Na by negative beta emission. Mass of 23Ne is 22.994465 amu mass of 23Na is 22.989768 amu. The maximum kinetic energy of emitted electrons neglecting the kinetic energy of recoiling product nucleus is ______ MeV.
पर्याय
4
3
2
1
MCQ
रिकाम्या जागा भरा
उत्तर
23Ne decays to 23Na by negative beta emission. Mass of 23Ne is 22.994465 amu mass of 23Na is 22.989768 amu. The maximum kinetic energy of emitted electrons neglecting the kinetic energy of recoiling product nucleus is 4 MeV.
Explanation:
\[\ce{_10^23Ne -> _11^23Na + e^- + \overline{v}}\]
Q = [m (23Ne) - m(23Na)] × 931.5 Me V
A = 4.375 Me V = 4.4 me V
Q ≃ 4 MeV
Q = KEy + KEe + E`overline"v"`
KEy is very very small
A ≈ KEe + E`overline"v"`
when KEe is maximum E`overline"v"` is negligible
KEe ≃ Q = 4 MeV
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