Question

A radioactive nucleus initially at rest decays by emitting an electron and neutron at right angles to one another. The momentum of the electron is $3.2 \times 10^{- 23}$ kgm/s and the momentum of neutron is 6.4 x $10^{- 23}$ kgm/sec. The direction of the recoiling nucleus with that of the electron motion is:

Moderate

Solution

$P_{e} = 3.2 \times 10^{- 23} kgm / s, P_{n} = 6.4 \times 10^{- 23} kgm / s$
$P_{R} = \sqrt{{P_{e}}^{2} + {P_{n}}^{2}} = \sqrt{{(3.2)}^{2} + {(6.4)}^{2}} \times 10^{- 23} kgm / s$
$\tan θ = \frac{P_{n}}{P_{e}} = \frac{6.4 \times 10^{- 23}}{3.2 \times 10^{- 23}} = 2$
According to law of conservation of momentum, the residual nucleus must move in a direction just opposite to that of $\vec{P_{R}}$ , as shown in
figure. Hence, direction of the recoiling nucleus with that of the electron motion $\emptyset$ = $π - θ = π - \tan^{- 1} (2) .$ .

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