Question

If an alpha particle were released from rest near the surface of a 257100fm nucleus, what would its kinetic energy be when far away?

Answer 1

The answer is 30.25MeV.

The radius of alpha particle is,

$\begin{aligned}\mathrm{r}_{a} &=r_{0} A^{1 / 3} \\&=\left(1.2 \times 10^{-15} \mathrm{~m}\right)(4)^{1 / 3} \\&=1.9 \times 10^{-15} \mathrm{~m}\end{aligned}$

Radius of ${ }_{100}^{257} \mathrm{Fm}$ is

$\begin{aligned}r_{F m} &=\left(1.2 \times 10^{-15} \mathrm{~m}\right)(257)^{1 / 3} \\&=7.62 \times 10^{-15} \mathrm{~m}\end{aligned}$

When the alpha particle is at the surface of mucleus, the seperation between alpha particle and nucleus is

$\begin{aligned}r &=r_{a}+r_{F m} \\&=1.9 \times 10^{-15} \mathrm{~m}+7.62 \times 10^{-15} \mathrm{~m} \\&=9.52 \times 10^{-15} \mathrm{~m}\end{aligned}$

potential energy is,

$\begin{aligned}U &=\frac{k(100 e)(2 e)}{r} \\&=\frac{\left(9 \times 10^{9}\right)(200)\left(1.6 \times 10^{-19} \mathrm{C}\right)^{2}}{9.52 \times 10^{-15} \mathrm{~m}}\end{aligned}$

$\begin{aligned}&=\left(4.84 \times 10^{-12} \mathrm{~J}\right)\left(\frac{1 \mathrm{MeV}}{1.6 \times 10^{-13} \mathrm{~J}}\right) \\&=30.25 \mathrm{MeV}\end{aligned}$

What is law of conservation of energy?

• According to law of conservation of energy, the kinetic energy of an alpha particle when far away is equal to the potnetial energy of alpha particle at surface of nucleus.
• The overall energy of an isolated system remains constant regardless of internal changes, with energy vanishing in one form and reappearing in another.

To learn more about conservation of energy visit:

https://brainly.com/question/2137260

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