determine the mass of a subatomic particle moving at 1.447×10^5 m/s that has e_k = 9.5367×10^(-21) j. be…

Explanation:

Step1: Recall kinetic - energy formula

The kinetic - energy formula is $E_{k}=\frac{1}{2}mv^{2}$, where $E_{k}$ is the kinetic energy, $m$ is the mass, and $v$ is the velocity. We need to solve for $m$.

Step2: Rearrange the formula for $m$

Starting from $E_{k}=\frac{1}{2}mv^{2}$, we can multiply both sides by 2 and then divide by $v^{2}$ to get $m = \frac{2E_{k}}{v^{2}}$.

Step3: Substitute the given values

We are given that $E_{k}=9.5367\times 10^{-21}\text{ J}$ and $v = 1.447\times 10^{5}\text{ m/s}$. Substitute these values into the formula: [ \begin{align*} m&=\frac{2\times(9.5367\times 10^{-21}\text{ J})}{(1.447\times 10^{5}\text{ m/s})^{2}}\ &=\frac{2\times9.5367\times 10^{-21}}{1.447^{2}\times 10^{10}}\ &=\frac{19.0734\times 10^{-21}}{2.093809\times 10^{10}}\ &=\frac{19.0734}{2.093809}\times10^{-21 - 10}\ &\approx9.11\times 10^{-31}\text{ kg} \end{align*} ]

Answer:

$9.11\times 10^{-31}\text{ kg}$