13.6 Write nuclear reaction equations for
(i) -decay of
(ii) -decay of
(iii) –-decay of
(iv) –-decay of
(v) +decay of
(vi) –-decay of
(vii) Electron capture of
13.6 Write nuclear reaction equations for
(i) -decay of
(ii) -decay of
(iii) –-decay of
(iv) –-decay of
(v) +decay of
(vi) –-decay of
(vii) Electron capture of
-
1 Answer
-
13.6 In - decay, there is a loss of 2 protons and 4 neutrons. In every decay, there is a loss of 1 proton and a neutrino is emitted from the nucleus. In every decay, there is a gain of 1 proton and an antineutrino is emitted from the nucleus.
+
+
+ +
+ +
+ +
+ +
+
Similar Questions for you
Q = [4 *4.0026 – 15.9994] *931.5 MeV
Q = 10.2 MeV
-(1)
for B,
for B,
-(2)
The reaction is X²? → Y¹²? + Z¹²?
Binding energies per nucleon are: X=7.6 MeV, Y=8.5 MeV, Z=8.5 MeV.
Gain in binding energy (Q) = (Binding energy of products) - (Binding energy of reactants)
Q = (120 × 8.5 + 120 × 8.5) - (240 × 7.6) MeV
Q = (2 × 120 × 8.5) - (240 × 7.6) MeV = 2040 - 1824 = 216 MeV.
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else
Sign Up on ShikshaOn Shiksha, get access to
- 65k Colleges
- 1.2k Exams
- 688k Reviews
- 1800k Answers