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 3.1×10−3 c, 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 × 108 m/s, k = 1/4πε 0 = 8.99 × 109 N · m2/C2, e = 1.60 × 10-19 C, m proton = 1.67 x 10-27 kg)

Answer 1

Answer:

$r = 5.32 \times 10^{-18} m$

Explanation:

When proton and alpha particle comes to closest distance of approach then the relative velocity of the two will become ZERO

so we will have

$m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v$

$4mv_o - mv_o = (4m + m) v$

$v = 0.6 v_o$

here we know that

$v_o = 3\times 10^{-3} (3 \times 10^8) = 9 \times 10^5 m/s$

$v = 5.4 \times 10^5 m/s$

now by energy conservation

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

$2.5mv_o^2 - 0.9 mv_o^2 = \frac{2ke^2}{r}$

$1.6\times (1.67 \times 10^{-27})(0.6 \times 3 \times 10^8)^2 = \frac{2(9\times 10^9)(1.6 \times 10^{-19})^2}{r}$

$8.65 \times 10^{-11} = \frac{4.61 \times 10^{-28}}{r}$

$r = 5.32 \times 10^{-18} m$