Question

A proton and art alpha particle are released from rest when they are 0.225 nm apart. The alpha particle (a helium nucleus) has essentially four times the mass and two times the charge of a proton. Find the maximum speed and maximum acceleration of each of these particles. When do these maxima occur: just following the release of the particles or after a very long time?

Answer 1

Answer:

Explanation

Acceleration will be maximum  at the moment when they are at the closest distance ie at the distance of .225 nm because at time,  force will be maximum.

Force at that time = k Qq/r² where q and Q are charges and r is distance between the two

Here q , charge on proton = 1.6 x 10⁻¹⁹

Q is charge on alpha particle = 2 x 1.6 x 10⁻¹⁹= 3.2 x 10⁻¹⁹

r = .225 x 10⁻⁹

Force =

Force F = $\frac{9\times10^9\times 1.6\times10^{-19}\times2\times1.6\times10^{-19}}{(.225\times10^{-9})^2}$

= 910.22 x 10⁻⁵ N

Acceleration in proton = force / mass

$\frac{910.22\times10^{-5}}{1.67\times10^ {-27}}$

=545.04 x 10²² ms⁻²

Acceleration in alpha particle will be 4 times less because its mass is 4 times more .

acceleration in alpha particle

= 136.25 x 10²²ms⁻²

Maximum speed will be when they reach at infinity . At that time all their energy ( potential ) will be converted into kinetic energy .

energy when they were at .225 nm

= $\frac{9\times 1.6\times3.2\times10^{-32}}{.225\times10^{-9}}$

=204.77 x 10⁻²³ J

This energy will be distributed among them in the inverse ratio of their mass

energy of proton = 204.77/5 x 4 x 10⁻²³ = 163.81 x 10⁻²³ J

energy of alpha particle = 204/5 x 1 x 10⁻²³

=40.8 x 10⁻²³ J.

Velocity of proton = $\sqrt{\frac{2\times E}{m} }$

=$\sqrt{\frac{2\times 163.81\times10^{-23}}{1.67\times10^{27}} }$

= 1400 m/s

velocity of alpha particle

$\sqrt{\frac{2\times 40.8\times10^{-23}}{4\times1.67\times10^{27}} }$

Velocity of alpha particle = 350 m/s