Question

A positron has a mass of 9.11 x 10^-31 kg, and charge qp = +e = +1.60 x 10^-19 C. It is moving towards an α particle (qα = +2e, mα = 6.66 x 10^-27 kg) with a speed of 3.00 x 10^6 m/s. At this instant the separation between the two is 2.00 x 10^-10 m. Assume α particle stays at rest. (a) Calculate the speed of positron at 1.00 x 10^-10 m from α particle. (b) What is the separation between the two when positron comes to rest?

Answer 1

Answer:

Part a)

$v = 2.25 \times 10^6 m/s$

Part b)

$r = 0.72 \times 10^{-10} m$

Explanation:

As we know that alpha particle remain at rest always so the energy of system of positron and alpha particle will remain constant

So we will have

$\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2$

here we know that

$q_1 = 1.6 \times 10^{-19} C$

$q_2 = 2(1.6 \times 10^{-19}) C$

also we have

$r_1 = 2.00 \times 10^{-10} m$

$r_2 = 1.00 \times 10^{-10} m$

$v_1 = 3.00 \times 10^6 m/s$

$m = 9.11 \times 10^{-31} kg$

now from above equation we have

$2.304 \times 10^{-18} + 4.0995\times 10^{-18} = 4.608 \times 10^{-18} + \frac{1}{2}(9.11 \times 10^{-31})v^2$

$v = 2.25 \times 10^6 m/s$

Part b)

Now when it come to rest then again by energy conservation we can say

$\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2$

now here final speed will be zero

$2.304 \times 10^{-18} + 4.0995\times 10^{-18} = \frac{(9/times 10^9)(1.6\times 10^{-19})(2\times 1.6 \times 10^{-19})}{r_2}$

$r = 0.72 \times 10^{-10} m$