Question

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

Answer 1

Answer:

a) ∆x∆v = 5.78*10^-5

   ∆v = 1157.08 m/s

b) 4.32*10^{-11}

Explanation:

To solve this problem you use the Heisenberg's uncertainty principle, that is given by:

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

where h is the Planck's constant (6.62*10^-34 J s).

If you assume that the mass of the electron is constant you have:

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

you use the value of the mass of an electron (9.61*10^-31 kg), and the uncertainty in the position of the electron (50nm), in order to calculate ∆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 the electron is a classical particle, the time it takes to traverse the channel is (by using the edge of the uncertainty in the velocity):

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