Question

The position of a neutron in a nucleus is known within an uncertainty of -5x10^-13 m. At what speeds might we expect to find it moving?

Short answer

The neutron moves at a speed of $∆v\ge 6.3\times {10}^6\frac{m}{s}$.

Step-by-step solution

Given data.

Mass of the neutron, m_n= 1.67x10^-27 kg .

Uncertainty in position, $∆x=5.0\times {10}^{-15}m$.

$\sout{h}=1.0545\times {10}^{-34}J\times s$.

$∆v$= ?

Heisenberg uncertainty principle.

The following equation describes the Heisenberg uncertainty principle for momentum.

$∆p∆x\ge \frac{\sout{h}}{2}\left(1\right)$

Equation for momentum.

The equation for momentum is as follows:

p = m v

By relating above equation with equation (1), then the following relationship will be:

$\begin{array}{rcl}∆p∆x& \ge & \frac{\sout{h}}{2}\\ m∆v∆x& \ge & \frac{\sout{h}}{2}\end{array}$

Expression for uncertainty in velocity.

The expression for uncertainty in velocity is,

$\begin{array}{rcl}m∆v∆x& \ge & \frac{\sout{h}}{2}\\ ∆v& \ge & \frac{\sout{h}}{2m∆x}\end{array}$

Estimation of velocity of neutron.

We can estimate the velocity of neutron by solving for$∆v$

$\begin{array}{rcl}∆v& \ge & \frac{\sout{h}}{2m∆x}\\ & \ge & \frac{1.0545\times {10}^{-34}}{2\times (1.67\times {10}^{-27})\times (5\times {10}^{-15})}\\ & \ge & 6.3\times {10}^6\frac{m}{s}\end{array}$

Thus, the neutron moves at a speed of $\begin{array}{rcl}∆v& \ge & 6.3\times {10}^6\frac{m}{s}\end{array}$.