Question

what is the speed of an electron with kinetic energy 880 ev? express your answer to two significant figures and include the appropriate units. part b what is the speed of an electron with kinetic energy 0.32 kev? express your answer to two significant figures and include the appropriate units.

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 ($m = 9.11\times10^{-31}\ kg$), and $v$ is the speed of the electron. We need to solve for $v$, so $v=\sqrt{\frac{2K}{m}}$. Also, we need to convert the energy from electron - volts (eV) to joules (J). We know that $1\ eV=1.6\times 10^{-19}\ J$.

Step2: Solve for part A

Given $K = 880\ eV$, first convert it to joules: $K = 880\times1.6\times10^{-19}\ J=1.408\times10^{-16}\ J$. Then, $v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2\times1.408\times 10^{-16}\ J}{9.11\times10^{-31}\ kg}}\approx1.77\times 10^{7}\ m/s$. Rounding to two significant figures, $v = 1.8\times 10^{7}\ m/s$.

Step3: Solve for part B

Given $K = 0.32\ keV=0.32\times10^{3}\ eV$, convert it to joules: $K = 0.32\times10^{3}\times1.6\times10^{-19}\ J = 5.12\times10^{-17}\ J$. Then, $v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2\times5.12\times10^{-17}\ J}{9.11\times10^{-31}\ kg}}\approx1.06\times 10^{7}\ m/s$. Rounding to two significant figures, $v = 1.1\times 10^{7}\ m/s$.

Answer:

Part A: $v = 1.8\times 10^{7}\ m/s$
Part B: $v = 1.1\times 10^{7}\ m/s$