Find the Q-value and the kinetic energy of the emitted α-particle in α-decay of (a) ${}_{88}^{226}{Ra} and (b)$ ${}_{86}^{220}{Rn}$
Given $m ({}_{88}^{226}{Ra}) = 226.02540 u; m ({}_{86}^{222}{Rn}) = 222.01750 u; m ({}_{86}^{220}{Rn}) = 220.01137 u; m ({}_{84}^{216}{Po}) = 216.00189 u .$

Solution

(a) The reaction invoved is,

${}_{88}^{226}{Ra} \to {}_{86}^{222}{Rn} + {}_{2}^{4}{He}$

The difference in mass between the original nucleus and the decay products = 226.02540 u - (222.01750 u + 4.00260 u)
= + 0.0053 u

$∴$ Energy equivalent or Q-value = 0.0053 x 931.5 MeV

= 4.93695 MeV

= 4.94 MeV

The decay products would emerge with total kinetic energy 4.94 MeV.

Momentum is conserved. If the parent nucleus is at rest, the daughter and the $α$ -particle have momenta of equal magnitude p but, in the opposite direction.

Kinetic energy, $K = \frac{p^{2}}{2 m} .$

Since, p is the same for the two particles therefore the kinetic energy divides inversely as their masses.

The $α$ -particle gets $\frac{222}{222 + 4}$ of the total i.e., $\frac{222}{226} \times 4.94 MeV or 4.85 MeV .$

(b) The difference in mass between the original nucleus and the decay products = 220.01137 u - (216.00189 u + 4.00260 u)
= 0.00688 u
$∴$ Q-value or Energy equivalent = 0.00688 x 931.5 MeV
= 6.41 MeV
Energy of the alpha particle, $E_{α} = \frac{216}{216 + 4} \times 6.41 MeV = 6.29 MeV .$

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Wave Optics

Find the Q-value and the kinetic energy of the emitted α-particle in α-decay of (a) ${}_{88}^{226}{Ra} and (b)$ ${}_{86}^{220}{Rn}$
Given $m ({}_{88}^{226}{Ra}) = 226.02540 u; m ({}_{86}^{222}{Rn}) = 222.01750 u; m ({}_{86}^{220}{Rn}) = 220.01137 u; m ({}_{84}^{216}{Po}) = 216.00189 u .$

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Wave Optics

Find the Q-value and the kinetic energy of the emitted α-particle in α-decay of (a) Ra88226 and (b) Rn86220Given mRa88226 = 226.02540 u; mRn86222 = 222.01750 u;mRn86220 = 220.01137 u; mPo84216 = 216.00189 u.

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Find the Q-value and the kinetic energy of the emitted α-particle in α-decay of (a) ${}_{88}^{226}{Ra} and (b)$ ${}_{86}^{220}{Rn}$
Given $m ({}_{88}^{226}{Ra}) = 226.02540 u; m ({}_{86}^{222}{Rn}) = 222.01750 u; m ({}_{86}^{220}{Rn}) = 220.01137 u; m ({}_{84}^{216}{Po}) = 216.00189 u .$