Consider a nucleus with a diameter of roughly meters.
Part A  

Consider a particle inside the nucleus. The uncertainty in its position is equal to the diameter of the nucleus. What is the uncertainty of its momentum? To find this, use . 
Express your answer in kilogrammeters per second to two significant figures.
ANSWER: 


Part B  

The uncertainty sets a lower bound on the average momentum of a particle in the nucleus. If a particle's average momentum were to fall below that point, then the uncertainty principle would be violated. Since the uncertainty principle is a fundamental law of physics, this cannot happen. Using kilogrammeters per second as the minimum momentum of a particle in the nucleus, find the minimum kinetic energy of the particle. Use kilograms as the mass of the particle. Note that since our calculations are so rough, this serves as the mass of a neutron or a proton. 
Express your answer in millions of electron volts to two significant figures.
ANSWER: 


Compare this to the normal energy scale for electrons in an atom, which is on the order of single electron volts. The characteristic energy scale for the nucleus seems to be roughly one million times that for electrons in an atom. This difference can be seen in calculations of the energy output per unit mass from coalburning power plants, which utilize chemical energy (energy associated with the electrons of an atom) compared to the energy output from nuclear reactors (power plants that harness the much higher energies of nuclei).
There is a mistake
ReplyDeleteThe right equation is ΔxΔp≥ℏ/2
So the right answer in part a) is two times smaller
Part A is actually 6.63*10^20
ReplyDeletep=(6.63*10^34/2)/5*10^15)=^