If an electron has a de Broglie wavelength of 0.250 nm, what is its momentum?

• A. 2.65 × 10^{-24} kg·m/s
• B. 6.63 × 10^{-34} kg·m/s
• C. 1.32 × 10^{-27} kg·m/s
• D. 9.11 × 10^{-31} kg·m/s

Answer: (A)

Verified step by step guidance

1. Recall the de Broglie wavelength formula that relates a particle's wavelength \(\lambda\) to its momentum \(p\): \[\lambda = \frac{h}{p}\] where \(h\) is Planck's constant and \(p\) is the momentum.
2. Rearrange the formula to solve for momentum \(p\): \[p = \frac{h}{\lambda}\]
3. Identify the given values: - Wavelength \(\lambda = 0.250\) nm (convert this to meters by multiplying by \$10^{-9}\() - Planck's constant \)h = 6.626 \times 10^{-34}$ J·s
4. Substitute the values into the momentum formula: \[p = \frac{6.626 \times 10^{-34}}{0.250 \times 10^{-9}}\]
5. Calculate the momentum \(p\) using the above expression to find the electron's momentum in kg·m/s.