Question

1. determine the de broglie wavelength, in nm, of an electron traveling at a speed of 5.93 × 10⁷ ms⁻¹. the mass of an electron is 9.11 × 10⁻³¹ kg. a. 8.88 × 10⁻¹ nm b. 3.35 × 10⁻⁴ nm c. 1.23 × 10⁻² nm d. 4.35 × 10¹ nm e. 3.14 × 10⁻³ nm

Explanation:

Step1: Recall de Broglie wavelength formula

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

Step2: Substitute the given values

Given \(m = 9.11\times10^{-31}\space kg\), \(v=5.93\times10^{7}\space m/s\), and \(h = 6.626\times10^{-34}\space J\cdot s\). Substitute these values into the formula:
\[
\lambda=\frac{6.626\times10^{-34}}{(9.11\times10^{-31})\times(5.93\times10^{7})}
\]

Step3: Calculate the denominator

First, calculate the product of \(m\) and \(v\):
\[
(9.11\times10^{-31})\times(5.93\times10^{7})=9.11\times5.93\times10^{-31 + 7}\approx9.11\times5.93\times10^{-24}
\]
\[
9.11\times5.93\approx53.92
\]
So, the denominator is approximately \(53.92\times10^{-24}=5.392\times10^{-23}\)

Step4: Calculate the wavelength in meters

Now, calculate \(\lambda\):
\[
\lambda=\frac{6.626\times10^{-34}}{5.392\times10^{-23}}\approx\frac{6.626}{5.392}\times10^{-34 + 23}\approx1.23\times10^{-11}\space m
\]

Step5: Convert meters to nanometers

Since \(1\space m = 10^{9}\space nm\), we convert \(\lambda\) to nanometers:
\[
\lambda=1.23\times10^{-11}\space m\times\frac{10^{9}\space nm}{1\space m}=1.23\times10^{-2}\space nm
\]

Answer:

C. \(1.23\times 10^{-2}\space nm\)