First slide

Third law

Question

A nuclide at rest emits an $α$ -particle. In this process:

Easy

A

$α$ - particle moves with large velocity and the nucleus remains at rest
B

both $α$ -particle and nucleus move with equal speed in opposite direction
C

both move in opposite directions but nucleus with greater speed
D

both move in opposite direction but $α$ -particle with greater speed.

Solution

As $F_{ext} = 0, hence \vec{P_{s}} = constant$
or $\vec{P_{α}} + \vec{P_{N}} = constant$
Because nuclide was at rest before emission of $α$ particle, hence according to law of conservation of momentum, $\vec{P_{α}} + \vec{P_{N}} = 0$
or $m_{α} \vec{v_{α}} + m_{N} \vec{v_{N}} = 0 or \vec{v_{α}} = - \frac{m_{N}}{m_{α}} \vec{v_{N}}$
Thus, $α$ -particle and the nucleus move in opposite directions. Further, as $m_{N} > > m_{α}, hence v_{α} > > v_{N}$

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