Question

A proton is confined within an atomic nucleus of diameter$4.0m$. Use a one-dimensional model to estimate the smallest range of speeds you might find for a proton in the nucleus.

Short answer

The average velocity is zero the velocity will be in the interval$-2.5\times {10}^{7}m/s\le v_x\le 2.5\times {10}^{7}m/s$. As the value of speed will not attain negative value, the uncertainty in the speed of neutron is $0m/s\le v_x\le 2.5\times {10}^{7}m/s$.

Step-by-step solution

Given Information

The classical deterministic approach was replaced by the probabilistic approach to describe the behavior of really tiny constitution of matter. These particles are called quantum particles and the mechanics that describes the physical behavior of these particles is called quantum physics.

Another feature of the quantum mechanics is the uncertainty involved in predicting the behavior of the wavelike particles. For example it is impossible to determine the precise location of the particle without losing information regarding its location. This uncertainty in position $\left(∆x\right)$and the momentum $\left(∆p_x\right)$is called Heisenberg uncertainty principle and is expressed through the following relation:

$∆x∆p_x\ge \frac{h}{2}$

The Planck's constant is $h$. Here we will estimate the uncertainty in the speed of proton confined in the atomic nucleus.

Expression

The uncertainty in the position of the nucleus confined in the nucleus is $∆x=4fm$. The uncertainty in momentum of the proton is expressed as follows:

role="math" $∆p_x=m∆v_x$

Substitute $∆p_x=m∆v_x$in the equation $∆x∆p_x\ge \frac{h}{2}$and simplify.

$$\begin{gathered} ∆xm∆v_x\ge \frac{h}{2} \\ ∆v_x\ge \frac{h}{2\left(∆xm\right)} \end{gathered}$$

Substitute $1.67\times {10}^{-27}kg$for $m,4\times {10}^{-15}m$for $∆x$, and $6.63\times {10}^{-34}J.s$for Planck's constant and solve.

$$\begin{gathered} ∆v_x\ge \frac{6.63\times {10}^{-34}J.s}{2\left(4\times {10}^{-15}m\right)\left(1.67\times {10}^{-27}kg\right)} \\ \ge 5\times {10}^{7}m/s \end{gathered}$$

As the average velocity is zero the velocity will be in the interval $-2.5\times {10}^{7}m/s\le v_x\le 2.5\times {10}^{7}m/s$. As the value of speed will not attain negative value, the uncertainty in the speed of neutron is$0m/s\le v_{x}2.5\times {10}^{7}m/s$.