Chemistry

Chemistry

Problem

Photons of wavelength 0.02 nm0.02\,\text{nm} are directed onto an electron in a microscope experiment. Assuming the position uncertainty is limited to Δx≈λ2\Delta x \approx \frac{\lambda}{2}, calculate the minimum possible uncertainty in the electron’s momentum. Take h=6.63×10−34 J sh = 6.63 \times 10^{-34}\,\text{J s}.
Intuitive explanation

This problem combines the wave nature of light used to observe the electron with Heisenberg's uncertainty principle. When you use short-wavelength photons to locate an electron precisely, you're limited by diffraction—the position uncertainty is roughly half the wavelength. The key insight is that this tight position constraint forces a large momentum uncertainty: the more precisely you know where the electron is, the less you can know about how fast it's moving. Use the uncertainty principle Δx⋅Δp≥h4π\Delta x \cdot \Delta p \geq \frac{h}{4\pi} to find the minimum momentum spread.

Intuitive explanation

This problem combines the wave nature of light used to observe the electron with Heisenberg's uncertainty principle. When you use short-wavelength photons to locate an electron precisely, you're limited by diffraction—the position uncertainty is roughly half the wavelength. The key insight is that this tight position constraint forces a large momentum uncertainty: the more precisely you know where the electron is, the less you can know about how fast it's moving. Use the uncertainty principle Δx⋅Δp≥h4π\Delta x \cdot \Delta p \geq \frac{h}{4\pi} to find the minimum momentum spread.

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