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  • Find the Q-value and the kinetic energy of the emitted alpha-particle in the alpha-decay of _88 ^ 226 Ra

Q.13.12  Find the Q-value and the kinetic energy of the emitted \alpha -particle in the \alpha-decay of

                  (a)\; _{88}^{226}\textrm{Ra}

                  Given m(_{88}^{226}\textrm{Ra})=226.02540\; u,      m(_{86}^{222}\textrm{Rn})=222.01750\; u,

                   m(_{86}^{222}\textrm{Rn})=220.01137\; u,     m(_{84}^{216}\textrm{Po})=216.00189\; u,

Answers (1)

best_answer

Mass defect is \Deltam

\Delta m=m(_{88}^{226}\textrm{Ra})-m(_{86}^{222}\textrm{Rn})-m(_{2}^{4}\textrm{He})

\Deltam=226.02540-222.0175-4.002603

\Deltam=0.005297 u

1 u = 931.5 MeV/c2

Q-value=\Deltam\times931.5

=4.934515 MeV

By using Linear Momentum Conservation and Energy Conservation 

Kinetic energy of alpha particle =

\frac{mass\ of\ nucleus\ after\ decay}{mass\ of\ nucleus\ before\ decay}\times Q-value

=\frac{222.01750}{226.0254}\times 4.934515

=4.847 MeV

Posted by

Sayak

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