219k views
0 votes
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?

User BDarley
by
7.3k points

1 Answer

4 votes

Answer:

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.

User Botenvouwer
by
7.1k points