As computer structures get smaller and smaller, quantum rules start to create difficulties. Suppose electrons move through a cha

Question

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As computer structures get smaller and smaller, quantum rules start to create difficulties. Suppose electrons move through a cha

nnel in a microprocessor. If we know that an electron is somewhere along the 50 nm length of the channel, what is ∆vx? If we treat the elec- tron as a classical particle moving at a speed at the outer edge of the uncertainty range, how long would it take to traverse the channel?

Physics

1 answer:

Softa [913] · Answer 1 · 2026-02-27T01:09:20+03:00

Answer:

a) ∆x∆v = 5.78*10^-5

∆v = 1157.08 m/s

b) 4.32*10^{-11}

Explanation:

This problem can be addressed using Heisenberg's uncertainty principle, which is expressed as:

$\Delta x\Delta p \geq \frac{\hbar}{2}$

Where h represents Planck’s constant (6.62*10^-34 J s).

Assuming that the electron's mass remains the same, we proceed as follows:

$\Delta x \Delta (m_ev)=m_e\Delta x\Delta v \geq \frac{\hbar}{2}$

Utilizing the electron's mass (9.61*10^-31 kg) and the uncertainty in position (50 nm), we can compute ∆x∆v and ∆v:

$\Delta x \Delta v\geq\frac{\hbar}{2m_e}=\frac{(1.055*10^{-34}Js)}{2(9.1*10^{-31}kg)}=5.78*10^{-5}\ m^2/s$

$\Delta v\geq\frac{5.78*10^{-5}}{50*10^{-9}m}=1157.08\frac{m}{s}$

If we treat the electron like a classic particle, the time required to cross the channel is determined using the upper limit of the uncertainty in velocity:

$t=\frac{x}{v}=\frac{50*10^{-9}m}{1157.08m/s}=4.32*10^{-11}s$