Suppose a { }_{88}^{226} Ra nucleus at rest a | vedclass

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(D) The $\alpha$-decay reaction is: ${ }_{88}^{226} Ra \longrightarrow { }_{86}^{222} Rn^* + { }_{2}^{4} \alpha$.The total energy released ($Q$-value)

Vedclass · Accepted Answer

$135$

Vedclass · Answer

(D) The $\alpha$-decay reaction is: ${ }_{88}^{226} Ra \longrightarrow { }_{86}^{222} Rn^* + { }_{2}^{4} \alpha$.The total energy released ($Q$-value) is given by: $Q = (M_{Ra} - M_{Rn} - M_{\alpha}) 	imes 931 	ext{ MeV}$.$Q = (226.005 - 222.000 - 4.000) 	imes 931 	ext{ MeV} = 0.005 	imes 931 	ext{ MeV} = 4.655 	ext{ MeV}$.Let $E_{\gamma}$ be the excitation energy of the ${ }_{86}^{222} Rn$ nucleus. The energy available for kinetic energy is $(Q - E_{\gamma})$.The kinetic energy of the $\alpha$ particle is $K_{\alpha} = \frac{A-4}{A} (Q - E_{\gamma})$,where $A = 226$.$4.44 	ext{ MeV} = \frac{222}{226} (4.655 - E_{\gamma})$.$4.655 - E_{\gamma} = 4.44 	imes \frac{226}{222} \approx 4.44 	imes 1.018 = 4.520 	ext{ MeV}$.$E_{\gamma} = 4.655 - 4.520 = 0.135 	ext{ MeV}$.Since $1 	ext{ MeV} = 1000 	ext{ keV}$,$E_{\gamma} = 0.135 	imes 1000 = 135 	ext{ keV}$.

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