Question

Assuming Heisenberg Uncertainty Principle to be true what could be the minimum uncertainty in de-Broglie wavelength of a moving electron accelerated by potential difference of 6 Volts whose uncertainty in position is n.m.

Answer 1

Answer on Question #59719, Physics / Quantum Mechanics

Assuming Heisenberg Uncertainty Principle to be true what could be the minimum uncertainty in de-Broglie wavelength of a moving electron accelerated by potential difference of 6 Volts whose uncertainty in position is n.m.

Find: $\Delta \lambda - ?$

Given:

$$U = 6 \, \mathrm{V}$$ $$m = 9,1 \times 10^{-31} \, \mathrm{kg}$$ $$h = 6,626 \times 10^{-34} \, \mathrm{J} \times \mathrm{s}$$ $$e = -1,6 \times 10^{-19} \, \mathrm{C}$$

Solution:

de-Broglie wavelength:

$$\lambda = \frac{h}{p} (1),$$

where $p$ – momentum of electron

We believe that the electron is a classic (“electron accelerated by potential difference of 6 Volts”).

The kinetic energy of the electron:

$$E = \frac{m v^2}{2} = \frac{m^2 v^2}{2 m} = \frac{p^2}{2 m} (2)$$ $$\text{Of } (2) \Rightarrow p = \sqrt{2 m E} (3)$$

Kinetic energy is numerically equal to the work. The work is done by the forces of electric field.

$$E = |e| \, U (4)$$

(4) in (3): $p = \sqrt{2 m |e| \, U} (5)$

Heisenberg Uncertainty Principle:

$$\Delta x \Delta p_x \geq \hbar (6),$$

where $\Delta x$ – uncertainties of coordinates,

$\Delta p_x$ – uncertainties of corresponding momentum’ projection,

$$\hbar = \frac{h}{2 \pi}$$ $$\text{Of } (1) \Rightarrow \Delta \lambda \Delta p \geq h (7),$$ $$\text{Of } (6) \Rightarrow 2 \pi \Delta x \Delta p_x \geq h (8)$$ $$\text{Of } (7) \text{ and } (8) \Rightarrow 2 \pi \Delta \lambda \Delta p \geq h (9)$$

We believe that $\Delta p \leq p$

(5) in (9): $2\pi \Delta \lambda \sqrt{2m|e|U} \geq h$ (10)

Of (10) $\Rightarrow \Delta \lambda = \frac{h}{2\pi\sqrt{2m|e|U}}$ (11)

Of (11) $\Rightarrow \Delta \lambda = 0,8 \times 10^{-10} \mathrm{~m}$

Answer:

$0,8 \times 10^{-10} \mathrm{~m}$

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