Question

A uranium-238 nucleus, initially at rest, emits an alpha particle with a speed of 1.4×10 raised to 7 m/s. Calculate the recoil speed of the residual nucleus thorium-234. Assume that the mass of a nucleus is proportional to the mass number.

Answer 1

ffffffffffffffffffffffffffffffffffffffffgggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggfffffffffffff

Answer 2

The recoil speed is  $2.39 \times 10^{5} \mathrm{m} / \mathrm{sec}$

Explanation:

The momentum in the initial(M )of mass number 238, that is to say, the uranium nucleus is zero as it is at rest. Once alpha particle gets released, the sum of moments of the alpha particle and remaining mass no 234 i.e. thorium is also zero because radioactive emissions are purely internal processes and are not influenced by any external force.  Accordingly,

As uranium 238 nucleus emits a $1.4 \times 10^{7} \mathrm{m} / \mathrm{sec}$ alpha particle. Let $v 2$ is considered as the thorium 234 residual nucleus velocity.

$m_{1} v_{1}=m_{2} v_{2}$

$4 \times 1.4 \times 10^{7}=234 \times v_{2}$

$\frac{4 \times 1.4 \times 10^{7}}{234}=v_{2}$

$v_{2}=0.0239 \times 10^{7}$

$v_{2}=2.39 \times 10^{5} \mathrm{M} / \mathrm{sec}$

As uranium 238 nucleus emits a $1.4 \times 10^{7} \mathrm{m} / \mathrm{sec}$ alpha particle. Let $v 2$ be the thorium 234 residual nucleus velocity.

$m_{1} v_{1}=m_{2} v_{2}$

$4 \times 1.4 \times 10^{7}=234 \times v_{2}$

$\frac{4 \times 1.4 \times 10^{7}}{234}=v_{2}$

$\frac{5.6 \times 10^{7}}{234}=v_{2}$

$v_{2}=0.0239 \times 10^{7}$

$v_{2}=2.39 \times 10^{5} \mathrm{m} / \mathrm{sec}$