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JEE Advanced · Physics · 28. Nuclear Physics

Suppose a Ra88226 nucleus at rest and in ground state undergoes α -decay to a Rn86222 nucleus in its excited state. The kinetic energy of the emitted α particle is found to be 4.44 MeV.Rn86222 nucleus then goes to its ground state by γ -decay. The energy of the emitted γ -photon is _______ keV,
[Given: atomic mass of  88226Ra=226.005u, atomic mass of  86222Rn=222.000u, atomic mass of α particle =4.000u,1u=931MeV/c2,c is speed of the light]

  1. A 130
  2. B 135
  3. C 140
  4. D 145
Verified Solution

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 \(=4.655-(4.44+0.08) \quad\)\(\left[E_\gamma=Q-\left(K E_\alpha+K E_{ Rn }\right)\right]\)
\(=0.135 MeV =135 KeV\)
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