Question

A proton and an 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 speedand maximum accelerationof each of these particles. When do these maxima occur: just following the release of the particles or after a very long time?

Short answer

The maximum speed of proton is $4.43\times {10}^4\mathrm{m}/\mathrm{s}$

The maximum acceleration of proton is $5.44\times {10}^{18}\mathrm{m}/{\mathrm{s}}^2$

The maximum speed of alpha particle is $1.11\times {10}^4\mathrm{m}/\mathrm{s}$

The maximum acceleration of the alpha particle is $1.36\times {10}^{18}\mathrm{m}/{\mathrm{s}}^2$

The maximum acceleration occurs immediately after the release. It takes a longer time to attain the maximum speed.

Step-by-step solution

(a) Determination of the maximum acceleration of proton and the alpha particle.

The proton charge is $q_p=+e$

The proton mass is $m_p=1.67\times {10}^{-27}\mathrm{kg}$

Right after the release, the force is maximum and therefore the acceleration is also maximum,

Acceleration $a_p=\frac{F}{m_0}$

Now, $F=k\frac{\left|q{q}^{\mathrm{\prime }}\right|}{r^2}$

And r is the distance which is equal to0.225 nm.

So, solve for F,

$\begin{array}{r}\mathrm{F}=\left(8.99\times {10}^9\mathrm{N}\times {\mathrm{m}}^2/{\mathrm{C}}^2\right)\frac{\left(2\right){\left(1.60\times {10}^{-19}\mathrm{C}\right)}^2}{{\left(0.225\times {10}^{-9}\mathrm{m}\right)}^2}\\ =9.09\times {10}^{-9}\mathrm{N}\end{array}$

$\begin{array}{r}{\mathrm{a}}_{\mathrm{p}}=\frac{9.09\times {10}^{-9}\mathrm{N}}{1.67\times {10}^{-27}\mathrm{kg}}\\ =5.44\times {10}^{18}\mathrm{m}/{\mathrm{s}}^2\end{array}$

Similarly, for the alpha particle,

The charge is $q_{\alpha }=+2e$

The mass is $m_{\alpha }=4m_p=6.68\times {10}^{-27}\mathrm{kg}$

Acceleration $a_{\alpha }=\frac{F}{m_{\alpha }}$

$\begin{array}{r}{\mathrm{a}}_{\alpha }=\frac{9.09\times {10}^{-9}\mathrm{N}}{6.68\times {10}^{-27}\mathrm{kg}}\\ =1.36\times {10}^{18}\mathrm{m}/{\mathrm{s}}^2\end{array}$

Thus the acceleration of the proton and alpha particle respectively is $5.44\times {10}^{18}\mathrm{m}/{\mathrm{s}}^2$ and $1.36\times {10}^{18}\mathrm{m}/{\mathrm{s}}^2$.

(b) Determination of the maximum speed of the alpha particle and the proton.

Speed is calculated by the law of conservation of energy,

${\mathrm{U}}_1+{\mathrm{K}}_1={\mathrm{U}}_2+{\mathrm{K}}_2$

Now, K₁ = 0 and U₂ = 0.

So, U₁ = K₂

Solve for U₁,

localid="1668229301464" $\begin{array}{r}{U}_1=\left(8.99\times {10}^9\mathrm{N}\times {\mathrm{m}}^2/{\mathrm{C}}^2\right)\frac{\left(2\right)\left(1.60\times {10}^{-19}\mathrm{C}\right)}{0.225\times {10}^{-9}\mathrm{m}}\\ =2.05\times {10}^{-18}\mathrm{J}\end{array}$

So this is the total kinetic energy of the system, localid="1668229340261" $\begin{array}{r}{K}_2=\frac{1}{2}{m}_p{v}_p^2+\frac{1}{2}{m}_{\alpha }{v}_{\alpha }^2\\ =2.05\times {10}^{-18}\mathrm{J}\end{array}$

This amount is divided between the proton and the alpha particle, Therefore, use the momentum conservation principle to solve for the speed of the proton.

$\begin{array}{r}{p}_1=p_2\\ m_p{v}_p-m_{\alpha }{v}_{\alpha }=0\\ v_a=\left(\frac{m_p}{m_{\alpha }}\right)v_p\end{array}$

So, kinetic energy is,

$\begin{array}{r}{K}_2=\frac{1}{2}{m}_p{\mathrm{v}}_p^2+\frac{1}{2}{m}_{\alpha }{\mathrm{v}}_{\alpha }^2\\ =\frac{1}{2}{m}_p{\mathrm{v}}_p^2\left(1+\frac{m_p}{m_{\alpha }}\right)\end{array}$

Solve for v_p,

localid="1668229807440" $\begin{array}{r}{V}_p=\sqrt{\frac{2K_2}{m_p\left(1+\left(\frac{m_p}{m_a}\right)\right)}}\\ =\sqrt{\frac{2\left(2.05\times {10}^{-18}\mathrm{J}\right)}{\left(1.67\times {10}^{-27}\mathrm{kg}\right)\left(1+\frac{1}{4}\right)}}\\ =4.43\times {10}^4\mathrm{m}/\mathrm{s}\end{array}$

Similarly, solve for v_α,

localid="1668229670474" $\begin{array}{r}{v}_a=\left(\frac{m_p}{m_a}\right)v_p\\ =\frac{1}{2}\frac{1}{2}\left(4.43\times {10}^4\mathrm{m}/\mathrm{s}\right)\\ =1.11\times {10}^4\mathrm{m}/\mathrm{s}\end{array}$

Thus the speed of the proton and alpha particle respectively is localid="1668229814402" $4.43\times {10}^4\mathrm{m}/\mathrm{s}$and $1.11\times {10}^4\mathrm{m}/\mathrm{s}$.