What is the uncertainty in its mass?

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The discussion revolves around calculating the uncertainty in the mass of a free neutron, given its mean lifetime of 900 seconds. Participants reference the Heisenberg uncertainty principle and Einstein's mass-energy equivalence, emphasizing the relationship between energy and time. The correct approach involves calculating the uncertainty in energy (ΔE) using the formula ΔEΔt ≥ h/4π, then converting this energy uncertainty into mass. One participant expresses confusion over the calculations but eventually arrives at a mass value, although they are reminded to check for rounding errors and unit consistency. The conversation highlights the importance of careful calculations in quantum mechanics.
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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?
 
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Hints: for exponential-decay processes, the uncertainty in the lifetime, \Delta t, equals the mean lifetime. Also, remember Einstein's mass-energy equivalence.
 
Okay... still do not see an equation. It is from our chapter on Quantum Mechanic of Atoms.
 
HINT: Heisenberg's uncertainty principle
 
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

? ? ?
 
The full expression is;

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

And jtbell gave you a bit hint

jtbell said:
Also, remember Einstein's mass-energy equivalence

~H
 
So I did it incorrectly? I am getting a bit confused...
 
Yes I know E = mc^2, but I still lost as to how to solve for uncertainty...
 
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
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
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
Opps. I mean kg! :-S
 
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