Question

The uncertainties in the velocities of two particles, A and B
are 0.05 and 0.02 ms⁻¹ respectively. The mass of B is five times
to that of the mass of A. What is the ratio of uncertainties
∆xA/∆xB in their positions ?
(a) 2 (b) 0.25 (c) 4 (d) 1

Answer 1

The ratio of uncertainties is Δx(A)  / Δx(B)  = 10 / 5 = 2

Option (A) is correct.

Explanation:

Using the relation,

Δx⋅Δv = h /  4πm

[Heisenberg' uncertainty principle]

Δx=   h / 4πm⋅Δv

Thus, Δx(A)  = h /  4π × 0.05 × m   ------ (1)

Δx (B ) = h /  4π×0.02×5 m  -------  (2)

Dividing (i) by (ii), we get

Δx(A)  / Δx(B)  = 0.02 x 5 / 0.05

Δx(A)  / Δx(B)  = 10 / 5 = 2

Thus the ratio of uncertainties is Δx(A)  / Δx(B)  = 10 / 5 = 2

Also learn more

Calculate the uncertainty in the position of an electron, if the uncertainty in its velocity is 5.7 × 10⁵ ms⁻¹. ?

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Answer 2

The ratio of uncertainties  in their positions ∆xA/∆xB = 2

Explanation:

According to Heisenberg's hypothesis it is not possible to find the position and momentum of tiny particles like an electron at an instant.

i.e. $\Delta x \cdot \Delta v=\frac{h}{4 \pi m}$

$\Rightarrow \Delta x=\frac{h}{4 \pi m \cdot \Delta v}$

It is given that, uncertainty in velocities of A and B are 0.05 and 0.02 ms⁻¹ respectively. Now by applying the above equation,

For the particle A,

$\Delta x_{A}=\frac{h}{4 \pi \times 0.05 \times m}$              

And, For the particle B,

$\Delta x_{B}=\frac{h}{4 \pi \times 0.02 \times 5 m}$

Therefore, $\frac{\Delta x_{A}}{\Delta x_{B}}=\frac{0.02 \times 5}{0.05}=\frac{10}{5}=2$

Hence the correct answer is option (a)