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* 10^7m/s`

Step-by-step 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* 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* 1.895* 10^5=4* v_2\\\\v_2=(238* 1.895* 10^5)/(4)=1.127* 10^7m/s`

Hence, the velocity of released alpha particle is
`1.127* 10^7m/s`