Question

2. An electron is confined to a potential well of width 10 cm. Calculate the minimum uncertainty
in its velocity.
Sol:​

Answer 1

Explanation:

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Answer 2

Answer:

The minimum uncertainty in the velocity, Δv = $5.77\times 10^{-4}m/s$.

Explanation:

Data given,

The width of the well, Δx = $10cm$ = $0.1m$

The minimum uncertainty in the velocity, Δv =?

From the Uncertainty principle, we can find out the minimum uncertainty in the velocity:

• $\Delta x \Delta p \ge \frac{h}{4\pi }$

Here,

• Δp = uncertainty in momentum = $m\Delta v$

• m = mass of electron = $9.1\times 10^{-31} kg$
• h = Planck's constant = $6.6\times 10^{-34} J-s$

Therefore, the equation becomes:

• $\Delta x m\Delta v \ge \frac{h}{4\pi }$
• $\Delta v \ge \frac{h}{4\pi m\Delta x}$
• $\Delta v \ge \frac{6.6\times 10^{-34} }{4\times 3.14 \times 9.1\times 10^{-31} \times 0.1}$
• $\Delta v \ge \frac{6.6\times 10^{-3} }{11.42}$
• $\Delta v \ge 5.77\times 10^{-4}m/s$

Hence, the minimum uncertainty in the velocity, Δv = $5.77\times 10^{-4}m/s$.