Question
Calculate the repulsive force on each proton in a helium nucleus separated in a vacuum by a distance of 2.50 fm . (Use $\mathrm{k}=9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2} ; 1 \mathrm{fm}=1 \times 10^{-15} \mathrm{~m}$; and $q_p=+1.602 \times 10^{-19} \mathrm{C}$.)
Step 1
In a helium nucleus, there are 2 protons. Each proton has a charge of \( q_p = +1.602 \times 10^{-19} \, \text{C} \). Show more…
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A helium nucleus has a charge of $+2 e$, and a neon nucleus has a charge of $+10 e$, where $e$ is the quantum of charge, $1.60 \times 10^{-19} \mathrm{C}$. Find the repulsive force exerted on one by the other when they are separated by a distance of $3.0$ nanometers $\left(1 \mathrm{~nm}=10^{-9} \mathrm{~m}\right)$. Assume the system to be in vacuum. Nuclei have radii of order $10^{-15} \mathrm{~m}$. We can assume them to be point charges in this case. Then $$ F_{E}=k_{0} \frac{q, q_{0}^{\prime}}{r^{2}}=\left(9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right) \frac{(2)(10)\left(1.6 \times 10^{-19} \mathrm{C}\right)^{2}}{\left(3.0 \times 10^{-9} \mathrm{~m}\right)^{2}}=5.1 \times 10^{-10} \mathrm{~N}=0.51 \mathrm{nN} $$
A helium nucleus has a charge of $+2 e$, and a neon nucleus has a charge of $+10 e$, where $e$ is the quantum of charge, $1.60 \times 10^{-19} \mathrm{C}$. Find the repulsive force exerted on one by the other when they are separated by a distance of $3.0$ nanometers $\left(1 \mathrm{~nm}=10^{-19} \mathrm{~m}\right)$. Assume the system to be in vacuum. Nuclei have radii of order $10^{-15} \mathrm{~m}$. We can assume them to be point charges in this case. Then $$ F_{E}=k_{0} \frac{q, q_{0}^{\prime}}{r^{2}}=\left(9,0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}, \frac{22(10)\left(1.6 \times 10^{-9} \mathrm{C}\right)^{2}}{\left(3.0 \times 10^{-9} \mathrm{~m}\right)^{2}}=5 \times 10^{-10} \mathrm{~N}=0.5 \mathrm{nN}\right. $$
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