Question

what is the velocity (in m/s) of an electron with a de broglie wavelength of 54.6 nm? round your answer to three significant figures. do not include units or use scientific notation.

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 of the particle. For an electron, $m = 9.11\times10^{-31}\ kg$. We can re - arrange the formula to solve for $v$: $v=\frac{h}{m\lambda}$.

Step2: Convert wavelength to SI units

The given wavelength $\lambda = 54.6\ nm=54.6\times10^{-9}\ m$.

Step3: Substitute values into the formula

Substitute $h = 6.63\times10^{-34}\ J\cdot s$, $m = 9.11\times10^{-31}\ kg$ and $\lambda = 54.6\times10^{-9}\ m$ into $v=\frac{h}{m\lambda}$.
\[

$$\begin{align*} v&=\frac{6.63\times 10^{-34}}{9.11\times 10^{-31}\times54.6\times 10^{-9}}\\ &=\frac{6.63\times 10^{-34}}{(9.11\times54.6)\times10^{-40}}\\ &=\frac{6.63\times 10^{-34}}{497.306\times10^{-40}}\\ &=\frac{6.63}{497.306}\times10^{6}\\ &\approx0.0133\times10^{6}\\ & = 1.33\times10^{4} \end{align*}$$

\]
Rounding to three significant figures, $v = 1.33\times10^{4}$.

Answer:

13300