# Homework Help: What is the uncertainty in its mass?

1. Apr 30, 2006

### jrd007

This is a really simple problem, but I cannot seem to find the equation. Any help? Thank you.

1) A free neutron has a mean life of 900 s. What is the uncertainty in its mass?

Last edited by a moderator: Jan 7, 2014
2. Apr 30, 2006

### Staff: Mentor

Hints: for exponential-decay processes, the uncertainty in the lifetime, $\Delta t$, equals the mean lifetime. Also, remember Einstein's mass-energy equivalence.

3. Apr 30, 2006

### jrd007

Okay... still do not see an equation. It is from our chapter on Quantum Mechanic of Atoms.

4. May 1, 2006

### Hootenanny

Staff Emeritus
HINT: Heisenberg's uncertainty principle

5. May 1, 2006

### jrd007

Yes, that would be:

(delta p)(delta x) > h or (energy)(time) > h....

So how does that help?

All I know is a time = 900 s.

So am I solving for energy? e =(1.06 x 10^-34 J)/(900 s) = 1.2 x 10^-37 J

? ? ?

6. May 1, 2006

### Hootenanny

Staff Emeritus
The full expression is;

$$\Delta E \Delta t \geq \frac{h}{4\pi}$$

And jtbell gave you a bit hint

~H

7. May 1, 2006

### jrd007

So I did it incorrectly? I am getting a bit confused...

8. May 1, 2006

### jrd007

Yes I know E = mc^2, but I still lost as to how to solve for uncertainty...

9. May 1, 2006

### Hootenanny

Staff Emeritus
Delta E is the uncertainty in energy of the neutron, you need to solve for delta E then convert this into mass using E = mc2.

~H

10. May 2, 2006

### jrd007

Btw, my book doe snot use 4 pie.

So I solve for E by : E = h/900 s = =1.06 x 10^-34 J/900 s? = 1.2 x 10^-37 J

Then I use E = mc^2
1.2 x 10^-37 J = m(3.0 x 10^8)^2
m = 1.54 x 10^-54 m

Is that correct! =)

11. May 2, 2006

### Hootenanny

Staff Emeritus
Your text uses a modified plank constant ( I think it is correct, I haven't checked it). Your working is right, but be careful, you are inducing rounding errors in you calculations. Also check your units of uncertainty in mass.

~H

12. May 2, 2006

### jrd007

Opps. I mean kg! :-S