Question

question 8
1 pts
what is the de broglie wavelength associated with an electron that is traveling at exactly 1/10th of the speed of light? the mass of an electron is 9.11 × 10^-31 kg
4.58 × 10^5 m
2.42 × 10^-11 m
2.18 × 10^5 m
7.02 × 10^-11 m

Explanation:

Step1: Determine the velocity of the electron

The speed of light \(c = 3\times10^{8}\ m/s\). The electron's speed \(v=\frac{1}{10}c=\frac{1}{10}\times3\times 10^{8}\ m/s = 3\times10^{7}\ m/s\)

Step2: Use the de - Broglie wavelength formula

The de - Broglie wavelength formula is \(\lambda=\frac{h}{mv}\), where \(h = 6.626\times10^{-34}\ J\cdot s\) (Planck's constant), \(m = 9.11\times10^{-31}\ kg\) (mass of the electron), and \(v\) is the velocity of the electron.

Substitute the values: \(\lambda=\frac{6.626\times 10^{-34}}{9.11\times10^{-31}\times3\times10^{7}}\)

First, calculate the denominator: \(9.11\times10^{-31}\times3\times10^{7}=2.733\times10^{-23}\)

Then, \(\lambda=\frac{6.626\times10^{-34}}{2.733\times10^{-23}}\)

Using the rule of exponents \(\frac{a^{m}}{a^{n}}=a^{m - n}\), we have \(\lambda=\frac{6.626}{2.733}\times10^{-34 + 23}\)

\(\frac{6.626}{2.733}\approx2.42\), and \(10^{-34 + 23}=10^{-11}\)

So, \(\lambda = 2.42\times10^{-11}\ m\)

Answer:

\(2.42\times 10^{-11}\ m\) (the second option)