TS EAMCET · Physics · Nuclear Physics
During the disintegration of a radioactive nucleus of mass number 208 at rest, two alpha particles each with kinetic energy E are emitted. The total kinetic energy of the emitted alpha particles and the daughter nucleus after the disintegration is
- A \(\frac{51 \mathrm{E}}{25}\)
- B \(\frac{51 \mathrm{E}}{50}\)
- C \(\frac{52 \mathrm{E}}{25}\)
- D \(\frac{26 \mathrm{E}}{25}\)
Answer & Solution
Correct Answer
(C) \(\frac{52 \mathrm{E}}{25}\)
Step-by-step Solution
Detailed explanation
\(A_D = 208 - 2(4) = 200\) \(m_D \approx 200\), \(m_\alpha \approx 4\) \(P_D = 2P_\alpha \Rightarrow K_D = \frac{P_D^2}{2m_D} = \frac{(2P_\alpha)^2}{2m_D} = \frac{4(2m_\alpha E)}{2m_D} = \frac{4m_\alpha E}{m_D}\) \(K_D = \frac{4(4)E}{200} = \frac{16E}{200} = \frac{2E}{25}\)…
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