The maximum potential energy due to electrost | vedclass

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(A) The electrostatic potential energy $U$ between two charges is given by $U = \frac{k q_1 q_2}{r}$.For two hydrogen nuclei (protons),the charge $q_1

Vedclass · Accepted Answer

$0.65$

Vedclass · Answer

(A) The electrostatic potential energy $U$ between two charges is given by $U = \frac{k q_1 q_2}{r}$.For two hydrogen nuclei (protons),the charge $q_1 = q_2 = 1.6 	imes 10^{-19} 	ext{ C}$.The distance of closest approach $d$ is equal to the sum of the radii of the two nuclei,so $d = 1.1 	ext{ fm} + 1.1 	ext{ fm} = 2.2 	imes 10^{-15} 	ext{ m}$.Substituting the values:$U = \frac{9 	imes 10^9 	imes (1.6 	imes 10^{-19})^2}{2.2 	imes 10^{-15}} 	ext{ J}$.$U = \frac{9 	imes 10^9 	imes 2.56 	imes 10^{-38}}{2.2 	imes 10^{-15}} 	ext{ J} \approx 1.047 	imes 10^{-13} 	ext{ J}$.To convert this to $	ext{MeV}$,divide by $1.6 	imes 10^{-13} 	ext{ J/MeV}$:$U = \frac{1.047 	imes 10^{-13}}{1.6 	imes 10^{-13}} 	ext{ MeV} \approx 0.65 	ext{ MeV}$.

The maximum potential energy due to electrostatic repulsion between two hydrogen nuclei is nearly (radius of the nucleus $= 1.1 \text{ fermi}$). $\left[\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ Nm}^2 \text{C}^{-2}\right]$ (in $MeV$)

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