Question

In a Rutherford scattering experiment, assume that an incident alpha particle (radius 1.80 m) is headed directly toward a target gold nucleus (radius 6.23fm).What energy must the alpha particle have to just barely “touch” the gold nucleus?

Short answer

The energy that the alpha particle must have to just barely “touch” the gold nucleus is 28.3 MeV .

Step-by-step solution

The given data

1. In a Rutherford scattering experiment, an alpha particle is targeted to the gold nucleus.
2. Radius of the alpha particle,${\mathrm{r}}_{\mathrm{a}}=1.80\mathrm{fm}$
3. Radius of the gold nucleus,${\mathrm{r}}_{\mathrm{Au}}=6.23\mathrm{fm}$

Understanding the concept of Rutherford scattering  

Rutherford scattering experiment involves high energy streams of α-particles being directed from a radioactive source at a thin sheet of gold to study the deflection caused to the α-particles. Here the problem refers to the closest distance of approach by the alpha particles to the gold sheet. Hence, considering that the kinetic energy of the particles gets converted into potential energy of the nucleus system, we can get the required value of energy possessed by an alpha particle.

Formula:

The electric potential energy between two charged bodies,$\mathbf{V}\mathbf{=}\frac{{\mathbf{kq}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{\mathbf{r}}$ (i)

Where, r is the separation between their centers or nuclei.

Calculation of the energy of the alpha particle

As per the concept, the kinetic energy of the alpha particle is same as the potential energy of the system having the both particles (K = U) for the condition of closest distance of approach by the alpha particles.

Now, we can get the charge of a particle from the concept that

q = Ze, where Z is the atomic number

Thus, charge of alpha particle, ${\mathrm{q}}_1=4\mathrm{e}$

Again for gold nucleus, ${\mathrm{q}}_2=79\mathrm{e}$

In order for theparticle to penetrate the gold nucleus, the separation between the centers of mass of the two particles must be no greater than

$$\begin{gathered} r=r_{Au}+r_{\alpha } \\ =6.23\mathrm{fm}+1.80\mathrm{fm} \\ =8.03\mathrm{fm} \end{gathered}$$

Thus, using the given data in equation (i), we can get the energy of the alpha particle as follows:

$$\begin{gathered} \mathrm{K}=\frac{\left(9\times {10}^9\mathrm{V}.\mathrm{m}/\mathrm{C}\right)\left(3\times 1.6\times {10}^{-19}\mathrm{C}\right)\left(79\mathrm{e}\right)}{\left(8.30\times {10}^{15}\mathrm{m}\right)} \\ =28.3\times {10}^6\mathrm{eV} \\ =28.3\mathrm{MeV} \end{gathered}$$

Hence, the value of the energy is 28.3 MeV .