question 18 of 20\nview policies\ncurrent attempt in progress\ndetermine the kinetic energy (in j/electron)…

Explanation:

Step1: Use de - Broglie wavelength formula

The de - Broglie wavelength formula is $\lambda=\frac{h}{p}$, where $\lambda$ is the wavelength, $h = 6.626\times10^{-34}\ J\cdot s$ is Planck's constant, and $p$ is the momentum. We can solve for $p$: $p=\frac{h}{\lambda}$. Given $\lambda = 4.25\times10^{-3}\ m$, then $p=\frac{6.626\times10^{-34}}{4.25\times10^{-3}}=\frac{6.626}{4.25}\times10^{-31}\ kg\cdot m/s\approx1.56\times10^{-31}\ kg\cdot m/s$.

Step2: Use kinetic energy - momentum relationship

The kinetic energy $K$ and momentum $p$ are related by the formula $K=\frac{p^{2}}{2m}$, where the mass of an electron $m = 9.11\times10^{-31}\ kg$. Substitute $p = 1.56\times10^{-31}\ kg\cdot m/s$ into the formula: $K=\frac{(1.56\times10^{-31})^{2}}{2\times9.11\times10^{-31}}=\frac{2.4336\times10^{-62}}{1.822\times10^{-30}}\ J$. $K\approx1.336\times10^{-32}\ J/electron$.

Answer:

$1.34\times 10^{-32}\ J/electron$ (rounded to three significant figures)