91Ó°ÊÓ

Two protons with kinetic energy are provided to us K=0.1 MeVfor each one, and the proton's charge is$\mathit{q}\mathbf{=}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathit{C}$.

The updated form of the energy principle entails the difference in energy of a system in its final state.

Write the relation between for energies and work done.

E_f = E_i + W …… (1)

Since, the contact between the two protons is incredibly rapid, the work done is zero.

The particle energy E is equal to the sum of each proton's kinetic energy and the system's potential energy.

The interaction between the two protons is the system here.

As a result, equation (I) would have the form

2K_f + U_f = 2K_i + U_i …… (2)

As a result, the two protons will eventually come to a halt, therefore, K_f = 0 Initially, they are opposed to one another.

Substitute U_i = 0 into equation (2).

$$\begin{gathered} 2K_i=U_f \\ 2K_i=\frac{1}{4\mathrm{π}{∈}_0}\frac{q_1{q}_2}{r}r \end{gathered}$$

$r=\frac{1}{4\mathrm{π}{∈}_0}\frac{q_1{q}_2}{2K_i}$ …… (3)

Here, $\frac{1}{4\mathrm{π}{∈}_0}=9×{10}^{9}N-m^7/C^2$

Putting the values into equation (3) to obtain r.

role="math" localid="1657531105055" $\begin{array}{rcl}r& =& \left(9×{10}^{9}N-m^2/C^2\right)\frac{{\left(1.6×{10}^{-19}C\right)}^2}{2\left(0.1×{10}^{6}eV\right)\left(1.6×{10}^{-19}J/eV\right)}\\ & =& 7.2×{10}^{-15}m\end{array}$

Thus, the separation between the protons when they finally come to a stop is $\begin{array}{rcl}r& =& 7.2×{10}^{-15}m\end{array}$.