Question

No object can travel faster than the speed of light, so it would appear evident that the uncertainty in the speed of any object is at most.

(a) What is the minimum uncertainty in the position of an electron, given that we know nothing about its speed except that it is slower than the speed of light?

(b) Repeat the calculation of part (a) for the position of a helium atom

Short answer

1. The uncertainty in the position of an electron is $1.93\times {10}^{-13}m$ .
2. The uncertainty in the position of the helium atom is $2.63\times {10}^{-17}m{s}^{-1}$.

Step-by-step solution

Heisenberg’s uncertainty principle

According to Heisenberg’s uncertainty principle, the uncertainty in position and velocity cannot be measured at the same time. This principle accounts for the dual particle nature of matter.

Calculation

a.

The uncertainty in position and velocity is calculated by the formula:

$\Delta x=\frac{h}{4\pi }\times \frac{1}{\Delta vm}$

*h is Planck’s constant, i.e.,$6.6.25\times {10}^{-34}Js$

*m is the mass of an electron, i.e., $9.1\times {10}^{-31}kg$

*v is the uncertainty in velocity, i.e.,$3\times {10}^{8}m{s}^{-1}$
.

*x is the uncertainty in the position.

On substituting the given values, we get

$\begin{array}{rcl}\Delta x& =& \frac{h}{4\pi }\times \frac{1}{\Delta vm}\\ & =& \frac{6.6.25\times {10}^{-34}Js}{4\times 3.14\times 3\times {10}^{8}m{s}^{-1}\times 9.1\times {10}^{-31}kg}\\ & =& 1.93\times {10}^{-13}m{s}^{-1}\\ & & \end{array}$

Thus, the uncertainty in the position of an electron is .

b.

The uncertainty in the position of the helium atom is calculated by using the formula:

*h is Planck’s constant, i.e., $6.6.25\times {10}^{-34}Js$

*m is the mass of helium atom, i.e.

*v is the uncertainty in velocity, i.e., .

*x is the uncertainty in the position.

On substituting the given values, we get

localid="1663782964808">Δx=h4π×1Δvm=6.6.25×10-34Js4×3.14×3×108ms-1×1.67×10-27kg=2.63×10-17ms-1

Thus, the uncertainty in position of the helium atom is $2.63\times {10}^{-17}m{s}^{-1}$.