Question

A proton inside a nucleus is confined to move inside the radius of the nucleus, about 2 fm (2 x 10-15 meters).

(a) what is the uncertainty in the momentum of the proton?

(b) what is the uncertainty in the velocity of the proton?

(c) for a proton whose velocity is equal to the uncertainty, how long does it take to cross the nucleus from one side to the other (a distance of 2r)?

(d) if the time to cross the nucleus is considered an uncertainty in the time, what is the uncertainty in the energy? Note: protons inside the nucleus of an atom typically have energy in the range of a few MeV – how does your answer compare to this?

Answer 1

The uncertainty in position =x = diameter of nucleus = 4fm

by Heisenberg uncertainty principle

xp = h/4

where p is the uncertainty in momentum

h = 6.63×10^-34Js

p = 6.63×10^-34 /(4)*4*10^-15

p = 0.1318 *10^-19 -----uncertainity in momentum -----answer for part a

b) momentum = mass *velocity

uncertainty in velocity = uncertainty in momentum/ m

v = p /m

mass of proton = 1.6726219 × 10^-27 Kg

v = 0.1318 *10^-19/ 1.6726219 × 10^-27

v= 0.07885 *10⁸ m/s----------------------------answer for part (b)

c) velocity =v

time = distance/ velocity

distance = 4fm

velocity = 0.07885 *10⁸ m/s

time to cross the nucleus = 4*10^-15 /0.0.07885 *10⁸

t = 50.72*10^-23S ------time required to cross nucleus ----answer or part (c)

d) Uncertainity principle can be also written as

Et = h/ 4

t = 50.72*10^-23S

E=  6.63×10^-34 /(4)*50.72*10^-23

= 0.0104 *10^-11J

E=0.104 pJ (pico joul) --------------------------anwer for part (d)

E = 0.624 MeV ---this is compaable for the proton energy inside the nucleus