Question

calculate the de broglie wavelength of an electron traveling at a speed of 3.15 x 10^2 m s^-1? the mass of an electron is 9.11 x 10^-28 g and 1 j = 1 kg m^2 s^-2.
2.31 x 10^-6 m
9.04 x 10^-26 m
2.89 x 10^-33 m
3.46 x 10^32 m
2.87 x 10^-20 m

Explanation:

Step1: Convert mass unit

First, convert the mass of the electron from grams to kilograms. Given the mass of an electron $m = 9.11\times10^{-28}\text{ g}$, and since $1\text{ kg}=1000\text{ g}$, then $m = 9.11\times 10^{-31}\text{ kg}$.

Step2: Recall de - Broglie wavelength formula

The de - Broglie wavelength formula is $\lambda=\frac{h}{p}$, where $h = 6.626\times10^{-34}\text{ J}\cdot\text{s}$ is Planck's constant and $p = mv$ is the momentum of the particle. Here, $v=3.15\times 10^{2}\text{ m/s}$.

Step3: Calculate momentum

Calculate the momentum $p$ of the electron: $p=mv=(9.11\times 10^{-31}\text{ kg})\times(3.15\times 10^{2}\text{ m/s}) = 2.86965\times10^{-28}\text{ kg}\cdot\text{m/s}$.

Step4: Calculate de - Broglie wavelength

Calculate the de - Broglie wavelength $\lambda$: $\lambda=\frac{h}{p}=\frac{6.626\times 10^{-34}\text{ J}\cdot\text{s}}{2.86965\times 10^{-28}\text{ kg}\cdot\text{m/s}}$. Since $1\text{ J}=1\text{ kg}\cdot\text{m}^{2}/\text{s}^{2}$, $\lambda=\frac{6.626\times 10^{-34}\text{ kg}\cdot\text{m}^{2}/\text{s}}{2.86965\times 10^{-28}\text{ kg}\cdot\text{m/s}}\approx 2.31\times 10^{-6}\text{ m}$.

Answer:

$2.31\times 10^{-6}\text{ m}$