# Problem: An electron has an uncertainty in its position of 552 pm. What is minimum possible value for the uncertainty in its velocity?a. 1.05 x 105 m/sb. 1.46 x 106 m/sc. 3.40 x 106 m/sd. 8.87 x 106 m/se. 2.21 x 107 m/s

###### FREE Expert Solution

We’re being asked to determine the uncertainty in the velocity of an electron with uncertainty in the position of 190 pm.

Recall that Heisenberg’s Uncertainty Principle states that we cannot accurately determine both the position and velocity of an electron. This means we can only know either one at any given time

Mathematically, this is expressed as:

$\overline{){\mathbf{\Delta x}}{\mathbf{·}}{\mathbf{\Delta p}}{\mathbf{\ge }}\frac{\mathbf{h}}{\mathbf{4}\mathbf{\pi }}}$

where:

h = Planck’s constant (6.626 × 10–34 kg • m2/s)

Δx = uncertainty in position (in m)

Δp = uncertainty in momentum (in kg • m/s)

Recall that momentum is expressed as:

$\mathbf{p}\mathbf{=}\mathbf{mv}$

where:

m = mass

v = velocity

Since the mass of an electron is constant, only its velocity can be uncertain

This gives us:

$\overline{){\mathbf{\Delta x}}{\mathbf{·}}{\mathbf{m\Delta v}}{\mathbf{\ge }}\frac{\mathbf{h}}{\mathbf{4}\mathbf{\pi }}}$

where:

m = mass (in kg)

Δv = uncertainty in velocity (in m/s)

88% (293 ratings) ###### Problem Details

An electron has an uncertainty in its position of 552 pm. What is minimum possible value for the uncertainty in its velocity?

a. 1.05 x 105 m/s
b. 1.46 x 10m/s
c. 3.40 x 106 m/s
d. 8.87 x 106 m/s
e. 2.21 x 107 m/s