JEE Advanced · Physics · 28. Nuclear Physics
Suppose a nucleus at rest and in ground state undergoes -decay to a nucleus in its excited state. The kinetic energy of the emitted particle is found to be nucleus then goes to its ground state by -decay. The energy of the emitted -photon is _______
[Given: atomic mass of atomic mass of atomic mass of particle is speed of the light]
- A 130
- B 135
- C 140
- D 145
Answer & Solution
Correct Answer
(B) 135
Step-by-step Solution
Detailed explanation
\(\Rightarrow \) Mass defect \(\Delta m=226.005-222.000-4.000 \quad\)\(\left[{ }_{88}^{226} R a \xrightarrow{\alpha-\text { decay }}{ }_{86}^{222} R n+{ }_2^4 H _{ e }+\gamma\right]\)
\(=0.005\text{ amu}\)
\(\therefore Q\) value \(=0.005 \times 931.5=4.655 MeV\)
Also \(\frac{K \cdot E_\alpha}{K \cdot E_{R n}}=\frac{m_{R n}}{m_\alpha}\)
\(\Rightarrow K \cdot E_{R n}=\frac{m_\alpha}{m_{R n}} \cdot K \cdot E_\alpha=\frac{4}{222}\) \(\times ~4.44=0.08 MeV\)
\(\therefore\) Energy of \(\gamma-\) Photon \(=4.655-(4.44+0.08) \quad\)\(\left[E_\gamma=Q-\left(K E_\alpha+K E_{ Rn }\right)\right]\)
\(=0.135 MeV =135 KeV\)
\(=0.005\text{ amu}\)
\(\therefore Q\) value \(=0.005 \times 931.5=4.655 MeV\)
Also \(\frac{K \cdot E_\alpha}{K \cdot E_{R n}}=\frac{m_{R n}}{m_\alpha}\)
\(\Rightarrow K \cdot E_{R n}=\frac{m_\alpha}{m_{R n}} \cdot K \cdot E_\alpha=\frac{4}{222}\) \(\times ~4.44=0.08 MeV\)
\(\therefore\) Energy of \(\gamma-\) Photon \(=4.655-(4.44+0.08) \quad\)\(\left[E_\gamma=Q-\left(K E_\alpha+K E_{ Rn }\right)\right]\)
\(=0.135 MeV =135 KeV\)
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