Question

Uncertainty in position of electron of 0.25gm is 10^-1. Uncertainty of its velocity is........ (Use Heinsberg's uncertainty formula)​

Answer 1

Answer:

$\displaystyle{\Delta v=2.108\times10^{-24} \ m/sec}$

Explanation:

According to  Heisenberg's uncertainty principle ,

$\displaystyle{\Delta x \times m\Delta v \geq }\dfrac{h}{4\pi}$

where , Δ x = Uncertainty in position

m = mass

Δ v = change in velocity

h = planck's constant

We have  Δ x = 10⁻¹

m = 0.25 g = 0.25 ×  10⁻³ kg

We have h = 6.626 × 10⁻³⁴ J sec.

We have to find Δ v

Put the value in formula we get

$\displaystyle{\Delta x \times m\Delta v \geq \dfrac{h}{4\pi}}\\\\\\\displaystyle{\Delta v \geq \dfrac{6.626\times10^{-34}}{4\pi\times 10^{-1} \times 0.25\times10^{-3}}}\\\\\\\displaystyle{\Delta v \geq \dfrac{6.626\times10^{-30}\times7}{4\times22\times0.25}}\\\\\\\displaystyle{\Delta v \geq \dfrac{6.626\times10^{-28}\times7}{4\times22\times25}}\\\\\\\displaystyle{\Delta v \geq \dfrac{46.38\times10^{-28}}{2200}}\\\\\\\displaystyle{\Delta v \geq \dfrac{46.38\times10^{-25}}{2.2}}$

$\displaystyle{\Delta v \geq 21.08\times10^{-25}}\\\\\\\displaystyle{\Delta v \geq 2.108\times10^{-24} \ m/sec}\\\\\\\displaystyle{\Delta v=2.108\times10^{-24} \ m/sec}$

Thus , we get answer .