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

Answer:

Distance of closest approach, $r=1.91\times 10^{-14}\ m$

Explanation:

It is given that,

Charge on proton, $q_p=e$

Charge on alpha particle, $q_a=2e$

Mass of proton, $m_p=1.67\times 10^{-27}\ kg$

Mass of alpha particle, $m_a=4m_p=6.68\times 10^{-27}\ kg$

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

$r=\dfrac{k2e^2(m_p+m_a)}{2m_am_pv_p^2}$

$r=\dfrac{9\times 10^9\times 2(1.6\times 10^{-19})^2(1.67\times 10^{-27}+6.68\times 10^{-27})}{2\times 6.68\times 10^{-27}\times 1.67\times 10^{-27}(0.01\times 3\times 10^8)^2}$

$r=1.91\times 10^{-14}\ m$

So, their distance of closest approach, as measured between their centers $1.91\times 10^{-14}\ m$. Hence, this is the required solution.