Question

calculate the de broglie wavelength of an electron traveling at a speed of 3.15 x 10^2 ms^-1? the mass of an electron is 9.11 x 10^-28 g and 1 j = 1 kg m^2 s^-2.

o 2.31 x 10^-6 m

o 9.04 x 10^-26 m

o 2.89 x 10^-33 m

o 3.46 x 10^32 m

o 2.87 x 10^-28 m

Explanation:

Step1: Convert mass to kg

The mass of electron $m = 9.11\times10^{-28}\ g$. Since $1\ g=10^{- 3}\ kg$, then $m = 9.11\times10^{-28}\times10^{-3}\ kg=9.11\times10^{-31}\ kg$.

Step2: Recall de - Broglie wavelength formula

The de - Broglie wavelength formula is $\lambda=\frac{h}{mv}$, where $h = 6.63\times10^{-34}\ J\cdot s$ (Planck's constant), $m$ is the mass of the particle and $v$ is the velocity of the particle. Here, $v = 3.15\times10^{2}\ m/s$.

Step3: Calculate the wavelength

Substitute the values of $h$, $m$ and $v$ into the formula:
\[

$$\begin{align*} \lambda&=\frac{h}{mv}\\ &=\frac{6.63\times 10^{-34}\ J\cdot s}{(9.11\times10^{-31}\ kg)\times(3.15\times10^{2}\ m/s)}\\ &=\frac{6.63\times 10^{-34}}{9.11\times3.15\times10^{-31 + 2}}\ m\\ &=\frac{6.63\times 10^{-34}}{28.6965\times10^{-29}}\ m\\ &=\frac{6.63}{28.6965}\times10^{-34 + 29}\ m\\ &\approx0.231\times10^{-5}\ m\\ &= 2.31\times10^{-6}\ m \end{align*}$$

\]

Answer:

A. $2.31\times10^{-6}\ m$