Question

An atom of uranium 238 emits an alpha particle (an atom of He) and recoils with a velocity of 1.895 * 10^ 5 m/sec . With velocity was the alpha particle released ?

Answer 1

Answer: The velocity of released alpha particle is $1.127\times 10^7m/s$

Explanation:

According to law of conservation of momentum, momentum can neither be created nor be destroyed until and unless, an external force is applied.

For a system:

$m_1v_1=m_2v_2$

where,

$m_1\text{ and }v_1$ = Initial mass and velocity

$m_2\text{ and }v_2$ = Final mass and velocity

We are given:

$m_1=238u\\v_1=1.895\times 10^{5}m/s\\m_2=4u\text{ (Mass of }\alpha \text{ -particle)}\\v_2=?m/s$

Putting values in above equation, we get:

$238\times 1.895\times 10^5=4\times v_2\\\\v_2=\frac{238\times 1.895\times 10^5}{4}=1.127\times 10^7m/s$

Hence, the velocity of released alpha particle is $1.127\times 10^7m/s$