Question

what is the velocity (in m/s) of an electron with a de broglie wavelength of 31 nm? round your answer to three significant figures. do not include units or use scientific notation. add your answer

Explanation:

Step1: Recall de - Broglie wavelength formula

The de - Broglie wavelength formula is $\lambda=\frac{h}{p}$, where $\lambda$ is the wavelength, $h = 6.63\times10^{-34}\ J\cdot s$ is Planck's constant and $p = mv$ is the momentum. We want to find $v$, and we can re - arrange the formula to $v=\frac{h}{m\lambda}$. The mass of an electron $m = 9.11\times10^{-31}\ kg$. The given wavelength $\lambda=31\ nm=31\times10^{-9}\ m$.

Step2: Substitute values into the formula

$v=\frac{6.63\times10^{-34}}{9.11\times10^{-31}\times31\times10^{-9}}$
$v=\frac{6.63\times10^{-34}}{(9.11\times31)\times10^{-40}}$
$v=\frac{6.63\times10^{-34}}{282.41\times10^{-40}}$
$v=\frac{6.63}{282.41}\times10^{6}$
$v\approx0.0235\times10^{6}=2.35\times 10^{4}$

Step3: Round to three significant figures

The value $2.35\times 10^{4}$ is already in three significant figures.

Answer:

$2.35\times 10^{4}$