Kinetic energy of the emitted alpha- particle | vedclass

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(D) The $\alpha-$ decay reaction is: ${}_{88}^{226}Ra \rightarrow {}_{86}^{222}Rn + {}_{2}^{4} \alpha$.First,calculate the mass defect $(\Delta m)$:$\

Vedclass · Accepted Answer

$4.85$

Vedclass · Answer

(D) The $\alpha-$ decay reaction is: ${}_{88}^{226}Ra ightarrow {}_{86}^{222}Rn + {}_{2}^{4} \alpha$.First,calculate the mass defect $(\Delta m)$:$\Delta m = m({}_{88}^{226}Ra) - [m({}_{86}^{222}Rn) + m({}_{2}^{4} \alpha)]$$\Delta m = 226.02540 - [222.01750 + 4.00260]$$\Delta m = 226.02540 - 226.02010 = 0.0053 \, u$.Next,calculate the $Q-$ value of the reaction:$Q = \Delta m 	imes 931.5 \, MeV/u$$Q = 0.0053 	imes 931.5 \approx 4.937 \, MeV$.The kinetic energy of the $\alpha-$ particle $(K_{\alpha})$ is given by the conservation of momentum and energy:$K_{\alpha} = \left( \frac{M_{daughter}}{M_{daughter} + M_{\alpha}} ight) Q$$K_{\alpha} = \left( \frac{222}{222 + 4} ight) 	imes 4.937 \, MeV$$K_{\alpha} = \frac{222}{226} 	imes 4.937 \approx 4.85 \, MeV$.

Kinetic energy of the emitted $\alpha-$ particle in the $\alpha-$ decay of ${}_{88}^{226}Ra$ will be,.......... $MeV$ (where $m_{\alpha} = 4.00260 \, u$,$m({}_{88}^{226}Ra) = 226.02540 \, u$ and $m({}_{86}^{222}Rn) = 222.01750 \, u$).

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