Question

The $U^{238}$ nucleus is unstable and decays into a $T{h}^{234}$ and an alpha particle. The alpha particle is emitted with a speed of $1.4\times {10}^{6}m/sec.$ The recoil speed of $T{h}^{234}$ is

• A. $\frac{4\times 1.4\times {10}^6}{234}m/s$
• B. $-\frac{4\times 1.4\times {10}^6}{234}m/s$
• C. $\frac{234}{4\times 1.4\times {10}^6}m/s$
• D. $0m/s$

Answer 1

Answer: (A)

$\frac{4\times 1.4\times {10}^6}{234}m/s$

Using momentum conservation:
$238\times 0=234\times v+4\times 1.4\times {10}^6$

$\Rightarrow v=-\left(\frac{4\times 1.4\times {10}^6}{234}\right)m/s$

Negative sign indicates that the thorium moves opposite to the alpha particle.