10. a. what is the binding energy released when an alpha particle of mass 6.64 x 10^(-27) kg escapes from…

Explanation:

Step1: Identify the formula

We use Einstein's mass - energy equivalence formula $E = mc^{2}$, where $E$ is energy, $m$ is mass and $c$ is the speed of light.

Step2: Solve for part a

Given $m = 6.64\times10^{-27}\text{ kg}$ and $c=3.00\times 10^{8}\text{ m/s}$. Substitute into the formula: $E = mc^{2}=(6.64\times 10^{-27}\text{ kg})\times(3.00\times 10^{8}\text{ m/s})^{2}$ $E=(6.64\times 10^{-27})\times(9.00\times 10^{16})\text{ J}$ $E = 5.976\times 10^{-10}\text{ J}$

Step3: Solve for part b

Given $m = 2.34\times10^{-28}\text{ kg}$ and $c = 3.00\times 10^{8}\text{ m/s}$. Substitute into the formula: $E=mc^{2}=(2.34\times 10^{-28}\text{ kg})\times(3.00\times 10^{8}\text{ m/s})^{2}$ $E=(2.34\times 10^{-28})\times(9.00\times 10^{16})\text{ J}$ $E = 2.106\times 10^{-11}\text{ J}$

Answer:

a. $5.976\times 10^{-10}\text{ J}$ b. $2.106\times 10^{-11}\text{ J}$