Question

use the kinetic energy given to determine the following. determine the velocity of an electron that has (e_{k}=6.11\times10^{-20} j). be sure your answer has the correct number of significant figures. note: reference the fundamental constants table for additional information.

Explanation:

Step1: Recall kinetic - energy formula

The kinetic - energy formula is $K = \frac{1}{2}mv^{2}$, where $K$ is the kinetic energy, $m$ is the mass of the electron ($m = 9.11\times10^{-31}\text{ kg}$ is the mass of an electron), and $v$ is the velocity. We need to solve for $v$.
First, re - arrange the formula for $v$: $v=\sqrt{\frac{2K}{m}}$.

Step2: Substitute the given values

Given $K = 6.11\times 10^{-20}\text{ J}$ and $m = 9.11\times10^{-31}\text{ kg}$.
$v=\sqrt{\frac{2\times(6.11\times 10^{-20}\text{ J})}{9.11\times10^{-31}\text{ kg}}}$.
First, calculate the value inside the square - root: $\frac{2\times6.11\times 10^{-20}}{9.11\times10^{-31}}=\frac{12.22\times 10^{-20}}{9.11\times10^{-31}}$.
Using the rule of exponents $\frac{a\times10^{n}}{b\times10^{m}}=\frac{a}{b}\times10^{n - m}$, we have $\frac{12.22}{9.11}\times10^{-20+31}\approx1.34\times10^{11}$.
Then $v=\sqrt{1.34\times10^{11}}$.
We can write $1.34\times10^{11}=13.4\times10^{10}$, and $\sqrt{1.34\times10^{11}}=\sqrt{13.4\times10^{10}}\approx3.66\times 10^{5}\text{ m/s}$.

Answer:

$3.66\times 10^{5}\text{ m/s}$