Question

Question #c0def

Answer 1

`Deltav >= 3.2 * 10^4` `"m s"^(-1)`

Explanation:

Start by writing down the equation for the de Broglie wavelength

`color(blue)(ul(color(black)(lamda_ "matter" = h/(m * v))))`

Here

• `lamda_ "matter"` is its de Broglie wavelength
• `h` is Planck's constant, equal to `6.626 * 10^(-34)"J s"`
• `m` is the mass of the particle
• `v` is its velocity

Now, the Heisenberg Uncertainty Principle states that it's impossible for us to measure both the position and the momentum of a particle with arbitrarily high precision.

Mathematically, this is expressed using the following inequality

`color(blue)(ul(color(black)(Deltax * Deltap >= h/(4pi))))`

Here

• `Deltax` is the uncertainty in position
• `Deltap` is the uncertainty in momentum
• `h` is Planck's constant

The uncertainty in momentum will depend on the mass of the particle, `m`, and on its uncertainty in velocity, `Deltav`

`color(blue)(ul(color(black)(Deltap = m * Deltav)))`

This means that you can rewrite the inequality that describes Heisenberg's Uncertainty Principle as

`Deltax * m* Deltav >= h/(4pi)" "color(darkorange)("(*)")`

Now, the problem tells you that this particle is moving with a velocity of `4 * 10^5` `"m s"^(-1)` and that its uncertainty in position is equal to its de Broglie wavelength.

This means that you have

`Deltax = lamda_"matter"`

or

`Deltax = h/(m * v)`

Plug this into `color(darkorange)("(*)")` to get

`color(red)(cancel(color(black)(h)))/(color(red)(cancel(color(black)(m))) * v) * color(red)(cancel(color(black)(m))) * Deltav >= color(red)(cancel(color(black)(h)))/(4pi)`

Since `v > 0`, you can multiply both sides by `v` without having to change the sign of the inequality, so

`Deltav >= v/(4pi)`

Plug in the value you have for the velocity of the particle to get

`Deltav >= (color(red)(cancel(color(black)(4))) * 10^5color(white)(.)"m s"^(-1))/(color(red)(cancel(color(black)(4))) * pi)`

`color(darkgreen)(ul(color(black)(Deltav >= 3.2 * 10^4color(white)(.)"m s"^(-1))))`

The answer is rounded to two sig figs, the number of sig figs you have for the velocity of the particle.