Question

The nucleus of $a^{125}Xe$atom (an isotope of the element xenon with mass $125u$) is $6.0fm$in diameter. It has $54$protons and charge $q=+54e.$

a. What is the electric force on a proton $2.0fm$from the surface of the nucleus?

b. What is the proton’s acceleration?

Hint: Treat the spherical nucleus as a point charge

Short answer

(a) The electric force of proton $F=497.6N$.

(b) The acceleration of the proton$a=2.99\times {10}^{29}\frac{\mathrm{m}}{{\mathrm{s}}^2}$.

Step-by-step solution

Find electric force (part a)

Because a proton is $2.0fm$from the SURFACE of the Xenon atom, you must add the $2.0fm$to the radius of the $Xe$atom, resulting in a distance of $5.0fm$between neutron and the centre of the $Xe$atom.

$2fm+3fm=5fm=5\times {10}^{-15}\mathrm{m}$

We must now determine the number of currently operates by the $Xe$atom. Because there are $54$protons in the problem, we must increase it by the charge of a proton, which is $1\times {10}^{-19}\mathrm{C}$.

$54\left(1\times {10}^{-19}\mathrm{C}\right)=8.64\times {10}^{-18}\mathrm{C}$

Substitute $F=\frac{k{q}_1{q}_2}{r^2}$in equation and$\mathrm{k}=9\times {10}^9\frac{N{m}^2}{C^2}$

$F=\frac{k{q}_1{q}_2}{r^2}$

$F=\frac{k\left(8.64\times {10}^{-18}C\right)\left(1.6{\times }^{-19}C\right)}{{\left(5\times {10}^{-15}m\right)}^2}$

$F=497.6N$

Find the acceleration of proton (part b)

To find the proton's force, use Newton's $F=ma$equation. In the equation, use the charge of a proton.

$F=m\times a$

$a=\frac{F}{m}$

$a=\frac{497.6\mathrm{N}}{1.661\times {10}^{-27}\mathrm{Kg}}$

$a=2.99\times {10}^{29}\frac{\mathrm{m}}{{\mathrm{s}}^2}$