Question

An effective treatment for some cancerous tumors involves irradiation with "fast" neutrons. The neutrons from one treatment source have an average velocity of 3.1 3 107 m/s. If the velocities of individual neutrons are known to within 2% of this value, what is the uncertainty in the position of one of them?

Answer 1

Answer:

$5.07\cdot 10^{-14} m$

Explanation:

The velocity of the neutrons is

$v=3.13\cdot 10^7 m/s$

The mass of a neutron is

$m=1.66\cdot 10^{-27} kg$

So their momentum is

$p=mv=(1.66\cdot 10^{-27})(3.13\cdot 10^7)=5.20\cdot 10^{-20}kg m/s$

The relative uncertainty on the velocity is 2 %. Assuming that the mass of the neutron is known with negligible uncertainty, then the relative uncertainty on the momentum of the neutron is equal to the relative uncertainty on the velocity, so 2%. Therefore, the absolute uncertainty on the momentum is

$\sigma_p = 0.02 p =0.02(5.20\cdot 10^{-20})=1.04\cdot 10^{-21} kg m/s$

Heisenber's uncertainty principle states that

$\sigma_x \sigma_p \geq \frac{h}{4\pi}$

where

$\sigma_x$ is the uncertainty on the position

h is the Planck constant

Solving for $\sigma_x$, we find the minimum uncertainty on the position:

$\sigma_x \geq \frac{h}{4\pi \sigma_p}=\frac{6.63\cdot 10^{-34}}{4\pi(1.04\cdot 10^{-21})}=5.07\cdot 10^{-14} m$