1. May 19, 2010

### SAT2400

1. The problem statement, all variables and given/known data
1. Since a charged pi meson at rest exists on average for only 26 ns, its energy cannot be measured with unlimited precision. Determine the minimum uncertainty in the meson's rest energy.
2. Determine the minimum uncertainty in energy of an atomic state if the state lasts for .10 us?

2. Relevant equations
delta(E) *delta(t) = h/2pi

3. The attempt at a solution
1. answer is 2x10^-27J..but I don't know how to get this answer.
2. answer is 5.3 x10^-28J..I don't know how to get this answer using the equation above.
I used h=6.63 x10^-34...but didn't get the right answers..

2. May 19, 2010

### diazona

Go back and check your references on the uncertainty principle; you have the equation wrong. It should be
$$\Delta E \Delta t = \frac{\hbar}{2}$$

3. May 19, 2010

### SAT2400

ok. but h=6.63x10^-34Js? right? 26ns should converted into what?

4. May 19, 2010

### mgb_phys

Again just make sure the units balance

energy (J) * time (s) = h_bar (Js) / 2
So you just need time in seconds, 26ns is 26x10^-9 s

Note that h_bar is h/2pi

5. May 19, 2010

### collinsmark

There should be an approximate sign in the above equation, not an equal sign.

$$\Delta E \Delta t \approx \hbar$$

There is a more formal equation for the minimum uncertainty (and it is slightly different than the above equation). It involves the the standard deviations of E and t (normally represented by $\sigma _E$ and $\sigma _t$ respectively), and it contains an 'equals' sign*. There is also a unit-less constant involved too, but you can look up the equation and you'll know what I'm talking about.

*The general form of the formal uncertainty equation involves a $$\geq$$ sign, but since we're talking about minimum uncertainty, it reduces to an = sign for this special case.

(By the way, $\hbar = h/2 \pi$.)

6. May 19, 2010

### SAT2400

2. answer is 5.3 x10^-28J..I don't know how to get this answer using the equation above. I used h=6.63 x10^-34Js and .1x10^-6s?...but didn't get the right answers..

7. May 19, 2010

### collinsmark

$$\sigma _E \sigma _t \geq \frac{\hbar}{2}$$
and $\hbar$ = 1.054571628×10-34 J·s.
So you know $\sigma _t$ and $\hbar$, so the minimum possible $\sigma _E$ must be....