How can I calculate the smallest speed of a proton in a nucleus?

[NoctusPartem]

Homework Statement

A proton is confined within an atomic nucleus of diameter 4.30 fm.
Estimate the smallest range of speeds you might find for a proton in the nucleus.

Relevant Equations

pL=hbar/2

My approach:

Assuming non-relativistic velocity:
$$p = mv$$

$$mvL = \frac{\bar{h}}{2}$$
$$v = \frac{\bar{h}}{2mL} = \frac{h}{4\pi mL}$$

$$v = (6.626*10^-34) / (4pi * 1.67*10^-27 * 4.3*10^-15)$$
$$v = 7.34*10^6 = .0245c$$

This answer is incorrect. What have I done wrong?

---

Edit: My lecturer's slides include the equation in the form:

I'm not sure if or why this form should be used instead.

 

[mitochan]

Hi.
I am not sure how incorrect your answer is. What is the right answer by your lecturer ?

 

[EM da Costa]

Hey.
I'm just a first year, so I haven't taken anything related to quantum.
But maybe you may find my attempt at this useful:

So by pythagoras, the Delta v in three dimensions should be this.
Assuming that the atom is uniform(all deltas are equal):

so according to this interpretation "n" should be related to the square root of the number of dimensions... But your lecturer's equation don't square "n" so I'm not sure what is wrong here.

 

[PeroK]

Homework Statement: A proton is confined within an atomic nucleus of diameter 4.30 fm.
Estimate the smallest range of speeds you might find for a proton in the nucleus.
Homework Equations: pL=hbar/2

My approach:

Assuming non-relativistic velocity:
$$p = mv$$

$$mvL = \frac{\bar{h}}{2}$$
$$v = \frac{\bar{h}}{2mL} = \frac{h}{4\pi mL}$$

$$v = (6.626*10^-34) / (4pi * 1.67*10^-27 * 4.3*10^-15)$$
$$v = 7.34*10^6 = .0245c$$

This answer is incorrect. What have I done wrong?

---

Edit: My lecturer's slides include the equation in the form:
View attachment 251429
I'm not sure if or why this form should be used instead.

I'm guessing that you are trying to use the Heisenberg Uncertainty Principle. But, since you don't say what you are trying to do, it's hard to know.

Meanwhile your lecturer seems to be using the lowest energy level for an infinite potential well.