Question

a nonrelativistic proton is confined to a length of 2.0 pm on the x-axis. what is the kinetic energy of the proton if its speed is equal to the minimum uncertainty possible in its speed

Answer 1

Answer:

K = 1.29eV

Explanation:

In order to calculate the kinetic energy of the proton you first take into account the uncertainty principle, which is given by:

$\Delta x \Delta p\geq \frac{h}{4\pi}$      (1)

Δx : uncertainty of position = 2.0pm = 2.0*10^-12m

Δp: uncertainty of momentum = ?

h: Planck's constant = 6.626*10^-34 J.s

You calculate the minimum possible value of Δp from the equation (1):

$\Delta p=\frac{h}{4\pi \Delta x}=\frac{6.626*10^{-34}J.s}{4\pi(2.0*10^{-12}m)}\\\\\Delta p=2.63*10^{-23}kg.\frac{m}{s}$

The minimum kinetic energy is calculated by using the following formula:

$k=\frac{(\Delta p)^2}{2m}$       (2)

m: mass of the proton = 1.67*10^{-27}kg

$k=\frac{(2.63*10^{-23}kgm/s)^2}{2(1.67*10^{-27}kg)}=2.08*10^{-19}J$

in eV you have:

$2.08*10^{-19}J*\frac{6.242*10^{18}eV}{1J}=1.29eV$

The kinetic energy of the proton is 1.29eV