Answer
See the explanation
Work Step by Step
To calculate the velocities of electrons with de Broglie wavelengths, we can use the de Broglie wavelength formula:
\[ \lambda = \frac{h}{mv} \]
Where:
- \( \lambda \) is the de Broglie wavelength,
- \( h \) is Planck's constant (\(6.62607015 \times 10^{-34} \, \text{m}^2 \, \text{kg/s}\)),
- \( m \) is the mass of the electron (\(9.10938356 \times 10^{-31} \, \text{kg}\)),
- \( v \) is the velocity of the electron.
Given the de Broglie wavelengths:
1. For \( \lambda = 1.0 \times 10^2 \, \text{nm} \):
\[ 1.0 \times 10^2 \, \text{nm} = \frac{h}{m \cdot v_1} \]
\[ v_1 = \frac{h}{m \cdot \lambda_1} \]
2. For \( \lambda = 1.0 \, \text{nm} \):
\[ 1.0 \, \text{nm} = \frac{h}{m \cdot v_2} \]
\[ v_2 = \frac{h}{m \cdot \lambda_2} \]
1. For \( \lambda = 1.0 \times 10^2 \, \text{nm} \):
\[ v_1 = \frac{{6.62607015 \times 10^{-34} \, \text{m}^2 \, \text{kg/s}}}{{9.10938356 \times 10^{-31} \, \text{kg} \times 1.0 \times 10^2 \times 10^{-9}}} \]
\[ v_1 \approx \frac{{6.62607015 \times 10^{-34}}}{{9.10938356 \times 10^{-38}}} \, \text{m/s} \]
\[ v_1 \approx 7.27 \times 10^3 \, \text{m/s} \]
2. For \( \lambda = 1.0 \, \text{nm} \):
\[ v_2 = \frac{{6.62607015 \times 10^{-34} \, \text{m}^2 \, \text{kg/s}}}{{9.10938356 \times 10^{-31} \, \text{kg} \times 1.0 \, \times 10^{-9}}} \]
\[ v_2 \approx \frac{{6.62607015 \times 10^{-34}}}{{9.10938356 \times 10^{-40}}} \, \text{m/s} \]
\[ v_2 \approx 7.27 \times 10^5 \, \text{m/s} \]
So, the velocities of electrons with de Broglie wavelengths of \( 1.0 \times 10^2 \, \text{nm} \) and \( 1.0 \, \text{nm} \) are approximately \( 7.27 \times 10^3 \, \text{m/s} \) and \( 7.27 \times 10^5 \, \text{m/s} \), respectively.
