Question

module 2b problem 17.14 an alpha particle (which is a helium nucleus, $q = + 2e$, $m=6.64\times10^{-27}\text{ kg}$) is emitted in a radioactive decay with $ke = 5.49\text{ mev}$. part a what is its speed? $v =$ m/s

Explanation:

Step1: Convert energy to joules

We know that $1\ MeV = 1.6\times10^{- 13}\ J$. So, $KE = 5.49\ MeV=5.49\times1.6\times10^{-13}\ J$.

Step2: Use kinetic - energy formula

The kinetic - energy formula is $KE=\frac{1}{2}mv^{2}$. We need to solve for $v$. Rearranging the formula gives $v=\sqrt{\frac{2KE}{m}}$. Substitute $KE = 5.49\times1.6\times10^{-13}\ J$ and $m = 6.64\times10^{-27}\ kg$ into the formula.
\[v=\sqrt{\frac{2\times5.49\times1.6\times10^{-13}}{6.64\times10^{-27}}}\]
\[v=\sqrt{\frac{17.568\times10^{-13}}{6.64\times10^{-27}}}\]
\[v=\sqrt{2.646\times10^{14}}\]
\[v = 1.627\times10^{7}\ m/s\]

Answer:

$1.63\times10^{7}\ m/s$ (rounded to three significant figures)