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

Answer 2

(a) The maximum speed of the proton is 5 x 10⁴ m/s and the maximum speed of the alpha particle is 2.5x 10⁴ m/s.

(b) The maximum acceleration of the proton is 5.45 x 10¹⁸ m/s² and the maximum acceleration of the alpha particle is 1.36 x 10¹⁸ m/s².

Force of attraction between the charges

The electrostatic force of attraction between the charges is calculated as follows;

$F = \frac{Kq_1q_2}{r^2} \\\\F = \frac{K(e)(2e)}{r^2} \\\\F = \frac{9\times 10^9(1.6 \times 10^{-19})(2\times 1.6\times 10^{-19})}{(0.225\times 10^{-9})^2} \\\\F = 9.1\times 10^{-9} \ N$

Acceleration of each particle

The acceleration of each particle is calculated as follows;

a = F/m

Acceleration of proton, with mass = 1.67 x 10⁻²⁷ kg

a = (9.1 x 10⁻⁹)/(1.67 x 10⁻²⁷)

a = 5.45 x 10¹⁸ m/s²

Acceleration of the alpha particle

a = F/4m

a = (9.1 x 10⁻⁹)/(4 x 1.67 x 10⁻²⁷)

a = 1.36 x 10¹⁸ m/s²

Potential difference due to the proton

$V= \frac{kq}{r} \\\\V = \frac{9\times 10^9 \times 1.6 \times 10^{-19}}{0.225 \times 10^{-9}} \\\\V = 6.4 \ V$

Potential energy

Work done in moving the alpha particle towards the proton is calculated as follows

$W = qV\\\\W = (2\times 1.6\times 10^{-19}) \times 6.4\\\\W = 2.05 \times 10^{-18} \ J$

Maximum speed will occur when the potential energy is converted into kinetic energy.

K.E = W

$\frac{1}{2} mv^2 = W\\\\v^2 = \frac{2W}{m} \\\\v = \sqrt{\frac{2W}{m} }$

Maximum speed of proton;

$v = \sqrt{\frac{ 2\times 2.05 \times 10^{-18}}{1.67 \times 10^{-27}} } \\\\v = 5 \times 10^4 \ m/s$

Maximum speed of the alpha particle

$v = \sqrt{\frac{ 2\times 2.05 \times 10^{-18}}{4(1.67 \times 10^{-27})} } \\\\v = 2.5 \times 10^4 \ m/s$

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