Question

An electron is allowed to move freely in cubic box of length of side =10cm. The uncertainty in its velocity will be

Answer 1

according to

Heisenberg's uncertainty principle ,

$m\Delta v\Delta x\leq\frac{h}{4\pi}$

here m is mass of electron, ∆v is uncertainty in velocity , ∆x is uncertainty in position and h is plank's constant.

m = 9.1 × 10^-31 Kg , ∆x = 10cm = 0.1m and h = 6.63 × 10^-34 Js

so, 9.1 × 10^-31 × ∆v × 0.1 ≤ 6.63 × 10^-34 Js/(4 × 3.14)

or, ∆v ≤ 6.63 × 10^-34/(4 × 3.14 × 9.1 × 10^-31 × 0.1)

or, ∆v ≤ 0.58 × 10^-3 m/s

hence, uncertainty in its velocity will be 5.8 × 10^-4 m/s

Answer 2

Answer: 3.34×10^-4

Explanation:

If a is side of cube, then ∆x= a√3

∆x= 10√3 cm = 10√3 × 10^-2m

∆v = h/4πm∆x

= 6.63×10^-34/4×3.14×9.1×10^-31×10√3×10^-2

= ∆v ≈ 3.34×10^-4 m/s