Respuesta :
Answer:
Part a)
[tex]v = 2.25 \times 10^6 m/s[/tex]
Part b)
[tex]r = 0.72 \times 10^{-10} m[/tex]
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
[tex]\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2[/tex]
here we know that
[tex]q_1 = 1.6 \times 10^{-19} C[/tex]
[tex]q_2 = 2(1.6 \times 10^{-19}) C[/tex]
also we have
[tex]r_1 = 2.00 \times 10^{-10} m[/tex]
[tex]r_2 = 1.00 \times 10^{-10} m[/tex]
[tex]v_1 = 3.00 \times 10^6 m/s[/tex]
[tex]m = 9.11 \times 10^{-31} kg[/tex]
now from above equation we have
[tex]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[/tex]
[tex]v = 2.25 \times 10^6 m/s[/tex]
Part b)
Now when it come to rest then again by energy conservation we can say
[tex]\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2[/tex]
now here final speed will be zero
[tex]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}[/tex]
[tex]r = 0.72 \times 10^{-10} m[/tex]
