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(A) The maximum potential energy occurs when the two hydrogen nuclei (protons) are in contact.The separation distance $r$ between the centers of the t

Vedclass · Accepted Answer

$0.65$

Vedclass · Answer

(A) The maximum potential energy occurs when the two hydrogen nuclei (protons) are in contact.The separation distance $r$ between the centers of the two nuclei is equal to the sum of their radii,which is $r = 2 	imes R = 2 	imes 1.1 	imes 10^{-15} \ m = 2.2 	imes 10^{-15} \ m$.The electrostatic potential energy $U$ is given by $U = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r}$.Substituting the values: $U = 9 	imes 10^9 	imes \frac{(1.6 	imes 10^{-19})^2}{2.2 	imes 10^{-15}} \ J$.$U = \frac{9 	imes 2.56 	imes 10^{-29} 	imes 10^9}{2.2 	imes 10^{-15}} = \frac{23.04 	imes 10^{-20}}{2.2 	imes 10^{-15}} \approx 10.47 	imes 10^{-5} \ J$.To convert to $MeV$,divide by $1.6 	imes 10^{-13} \ J/MeV$:$U = \frac{10.47 	imes 10^{-5}}{1.6 	imes 10^{-13}} \approx 6.54 	imes 10^7 \ eV = 0.65 \ MeV$.

The maximum potential energy due to electrostatic repulsion between two hydrogen nuclei is nearly (radius of the nucleus $= 1.1 \ Fermi$). $\left[ \frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \ N \ m^2 \ C^{-2} \right]$ (in $MeV$)

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