Question

An alpha particle is a nucleus of Helium. It has twice the charge and four times the mass of the proton. When they were very far away from each other, but headed toward directly each other, a proton and an alpha particle each had an initial speed of 8.5e-3c,where c is the speed of light. What is their distance of closest approach? There are two conserved quantities. Make use of both of them. (c = 3.00

Answer 1

Answer:

$r = 2.65 \times 10^{-14} m$

Explanation:

By energy conservation we know that initially when both are far apart from each other then in that case total electrostatic kinetic energy of two charges will convert into electrostatic potential energy of two charges

so we have

$\frac{1}{2}m_1v_1^2 + \frac{1}{2}m_1v_2^2 = \frac{1}{2}(m_1 + m_2)v^2 + \frac{kq_1q_2}{r}$

now we know that there is no external force on this system so momentum is conserved here

$m_1v_1 - m_2v_2 = (m_1 + m_2) v$

$4mv - mv = 5m v_f$

$v_f = 0.6 v$

now plug in all values in first equation

$\frac{1}{2}4mv^2 + \frac{1}{2}mv^2 = \frac{1}{2}(4m + m)(0.6v)^2 + \frac{k(e)(2e)}{r}$

$1.6 mv^2 = \frac{2ke^2}{r}$

$r = \frac{2ke^2}{1.6 mv^2}$

$r = \frac{2(9\times 10^9)(1.6\times 10^{-19})^2}{1.6(1.67 \times 10^{-27})(8.5\times 10^{-3})^2(3\times 10^8)^2}$

$r = 2.65 \times 10^{-14} m$