Question

A proton and an alpha particle (q = +2e, m = 4 u) are fired directly toward each other from far away, each with an initial speed of 0.010c. What is their distance of closest approach, as measured between their centers?

Answer 1

Distance of closest approach,
`r=1.91* 10^(-14)\ m`

Step-by-step explanation:

It is given that,

Charge on proton,
`q_p=e`

Charge on alpha particle,
`q_a=2e`

Mass of proton,
`m_p=1.67* 10^(-27)\ kg`

Mass of alpha particle,
`m_a=4m_p=6.68* 10^(-27)\ kg`

The distance of closest approach for two charged particle is given by :

`r=(k2e^2(m_p+m_a))/(2m_am_pv_p^2)`

`r=(9* 10^9* 2(1.6* 10^(-19))^2(1.67* 10^(-27)+6.68* 10^(-27)))/(2* 6.68* 10^(-27)* 1.67* 10^(-27)(0.01* 3* 10^8)^2)`

`r=1.91* 10^(-14)\ m`

So, their distance of closest approach, as measured between their centers
`1.91* 10^(-14)\ m`. Hence, this is the required solution.