Nuclear Physics And Radioactivity Question 460
Question: When the nucleus of a radium-226, which is at rest, decays, an $ \alpha $ particle and the nucleus of radon are created. The released energy during the decay is 4.87 MeV, which appears as the kinetic energy of the two resulted particles. Calculate the kinetic energy of a particle and radon nucleus.
Options:
A) 4.78 MeV and 0.09 MeV
B) 4.67 MeV and 0.2 MeV
C) 4.84 MeV and 0.03 MeV
D) 4.81 MeV and 0.06 MeV
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ {k_{\alpha }}=\frac{M}{M+4}Q=\frac{222}{226}\times 4.87=4.87MeV $
BETA
