Question

a.) What is the minimum uncertainty in an electron's velocity (Δmin) if the position is known within 13 Å? b.) What is the minimum uncertainty in a helium atom's velocity (Δmin) if the position is known within 1.0 Å?

Answer 1

Answer :

(a) The minimum uncertainty in an electron's velocity is, $4.5\times 10^{4}m/s$

(b) The minimum uncertainty in a helium atom's velocity is, $7.9\times 10^{1}m/s$

Explanation :

According to the Heisenberg's uncertainty principle,

$\Delta x\times \Delta p=\frac{h}{4\pi}$ ...........(1)

where,

$\Delta x$ = uncertainty in position

$\Delta p$ = uncertainty in momentum

h = Planck's constant

And as we know that the momentum is the product of mass and velocity of an object.

$p=m\times v$

or,

$\Delta p=m\times \Delta v$      .......(2)

Equating 1 and 2, we get:

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

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

(a) Given:

m = mass of electron = $9.11\times 10^{-31}kg$

h = Planck's constant = $6.626\times 10^{-34}Js$

$\Delta x$ = $13\AA=13\times 10^{-10}m$

conversion used : $(1\AA=10^{-10}m)$

Now put all the given values in the above formula, we get:

$\Delta v=\frac{6.626\times 10^{-34}Js}{4\times 3.14\times (13\times 10^{-10}m)\times (9.1\times 10^{-31}kg)}$

$\Delta v=4.5\times 10^{4}m/s$

The minimum uncertainty in an electron's velocity is, $4.5\times 10^{4}m/s$

(b) Given:

m = mass of helium atom = $6.646\times 10^{-27}kg$

h = Planck's constant = $6.626\times 10^{-34}Js$

$\Delta x$ = $1.0\AA=1.0\times 10^{-10}m$

conversion used : $(1\AA=10^{-10}m)$

Now put all the given values in the above formula, we get:

$\Delta v=\frac{6.626\times 10^{-34}Js}{4\times 3.14\times (1.0\times 10^{-10}m)\times (6.646\times 10^{-27}kg)}$

$\Delta v=7.9\times 10^{1}m/s$

The minimum uncertainty in a helium atom's velocity is, $7.9\times 10^{1}m/s$