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The nucleus of an atom is ${}_{92}^{235}Y$ initially at rest, decays by emitting an $α$ particle as per the equation
${}_{92}^{235}Y \to {}_{90}^{231}X + {}_{2}^{4}H e + E n e r g y$
It is given that the binding energies per nucleon of the parent and the daughter nuclei are $7.8 M e V$ and $7.835 M e V$ respectively and thar of $α$ particle is $7.07 M e V / n u c l e o n$ . Assuming the daughter nucleous to be formed in the unexcited state and neglecting its share in the energy of the reaction, calculate the speed of the emitted $α$ particle. Take mass of $α$ particle to be $6.68 \times 10^{- 27} k g$

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$ν_{α} = 1.57 \times 10^{6} m s^{- 1}$

$ν_{α} = 3.57 \times 10^{7} m s^{- 1}$

$ν_{α} = 1.57 \times 10^{5} m s^{- 1}$

$ν_{α} = 1.57 \times 10^{7} m s^{- 1}$

answer is D.

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Detailed Solution

$Q = [(7.835 \times 231) + (7.07 \times 4) - (7.8 \times 235)] M e V \Rightarrow Q = 5.18 M e V \Rightarrow Q = 5.18 \times 1.6 \times 10^{- 13} J$

This entire kinetic energy is taken by the $α$ particle so

$\frac{1}{2} m_{α} {ν_{α}}^{2} = 5.18 \times 1.6 \times 10^{- 13} \Rightarrow \frac{1}{2} \times 6.68 \times 10^{- 27} {ν_{α}}^{2} = 5.18 \times 1.6 \times 10^{- 13} \Rightarrow ν_{α} = 1.57 \times 10^{7} m s^{- 1}$

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