Question

The oxygen nucleus ¹⁶O has a radius of 3.0 fm.

a. With what speed must a proton be fired toward an oxygen nucleus to have a turning point 1.0 fm from the surface? Assume the nucleus remains at rest.

b. What is the proton’s kinetic energy in MeV?

Short answer

a) Speed of the proton is$2.35\times {10}^{7}m/s$

b) Kinetic energy = 2.88 MeV

Step-by-step solution

Step 1. Given information is :Radius of nucleus of Oxygen = 3.0 fmAtomic mass of Oxygen = 16u

We need to find out :

a) Speed with which the proton be fired toward an oxygen nucleus to have a turning point 1.0 fm from the surface.

b) Kinetic energy of proton in MeV.

Step 2. Using conservation of energy.

$$\begin{gathered} K_{initial}+U_{initial}=K_{final}+U_{final} \\ {U}_{initial}=0 \\ {K}_{final}=0 \\ {U}_{final}=\frac{1}{4\mathrm{\pi }{\in }_{\mathrm{o}}}\frac{q_1{q}_2}{r+d} \\ {U}_{final}=\frac{1}{4\mathrm{\pi }{\in }_{\mathrm{o}}}\frac{\left(e\right)\left(8e\right)}{r+d} \\ {K}_{initial}+0=0+\frac{1}{4\mathrm{\pi }{\in }_{\mathrm{o}}}\frac{\left(e\right)\left(8e\right)}{r+d} \\ {K}_{initial}=\frac{1}{4\mathrm{\pi }{\in }_{\mathrm{o}}}\frac{\left(e\right)\left(8e\right)}{r+d} \\ \frac{m{v_i}^2}{2}=\frac{1}{4\mathrm{\pi }{\in }_{\mathrm{o}}}\frac{\left(e\right)\left(8e\right)}{r+d} \\ {v}_i=\sqrt{\frac{2}{4\mathrm{\pi }{\in }_{\mathrm{o}}m}\frac{\left(e\right)\left(8e\right)}{r+d}} \\ {v}_i=\sqrt{\frac{2}{4\mathrm{\pi }{\in }_{\mathrm{o}}\times 1.67\times {10}^{-27}}\frac{(1.6\times {10}^{-19})(8\times 1.6\times {10}^{-19})}{3\times {10}^{-15}+1.0\times {10}^{-15}}} \\ {v}_i=2.35\times {10}^{7}m/s \end{gathered}$$

For Kinetic Energy :

$$\begin{gathered} K_{iniitial}=\frac{m{v_i}^2}{2} \\ {K}_{iniitial}=\frac{1.67\times {10}^{-27}kg\times {\left(2.35\times {10}^{7}m/s\right)}^2}{2} \\ {K}_{iniitial}=4.614\times {10}^{-13}J=2.88MeV \end{gathered}$$