Calculating the Lifetime of J/ψ Meson Using the Uncertainty Principle

[airkapp]

I'm just being introduced to particle physics and wondering if I'm going about this correct..can someone help me? thanks

The measured width of the J/ψ meson is 88 keV. Estimate its lifetime.

Lifetime (s) of J/ψ meson is 8E-21 s
Using uncertainty principle: ΔE = h/2πΔt isolate for delta t?
8E-21s – 5E-22s
= 7.5E-21 s ?

 

[dextercioby]

Yes,u have to use the uncertainty time-energy...It should be ~$7\cdot 10^{-21}s$

Daniel.

 

[airkapp]

Yes,u have to use the uncertainty time-energy...It should be ~$7\cdot 10^{-21}s$

Daniel.

thanks Daniel,

sorry to bother you again but do you think you can detail that out for me a little more..I think I'm having a little trouble following the steps of proportion to get $7\cdot 10^{-21}s$

thanks

 

[dextercioby]

Okay.Assume the general uncertainty relation being satisfied in the equality limit also for time & energy (nice discussion on this relation,chapter 2 of [1]).

$$\Delta E\Delta t=\frac{\hbar}{2}$$

Therefore

$$\Delta t=\frac{\hbar}{2\Delta E}\simeq\frac{6.626\cdot 10^{-34}\mbox{Js}}{4\pi\cdot 88\cdot 10^{3}\cdot 1.6\cdot 10^{-19}\mbox{J}}\simeq 3.74\cdot 10^{-21} \mbox{s}$$

The difference from the prior result (post #2) comes from using $=\frac{\hbar}{2}$ instead of $\approx\hbar$ which is cusomery as well.


Daniel.


------------------------------------------------------
[1] J.J.Sakurai,"Modern Quantum Mechanics",any of the 2 editions.

 

[dextercioby]

Hmmmm,something weird here. The above message appears as unposted. Can anyone see it...?

Daniel.

 

[airkapp]

Okay.Assume the general uncertainty relation being satisfied in the equality limit also for time & energy (nice discussion on this relation,chapter 2 of [1]).

$$\Delta E\Delta t=\frac{\hbar}{2}$$

Therefore

$$\Delta t=\frac{\hbar}{2\Delta E}\simeq\frac{6.626\cdot 10^{-34}\mbox{Js}}{4\pi\cdot 88\cdot 10^{3}\cdot 1.6\cdot 10^{-19}\mbox{J}}\simeq 3.74\cdot 10^{-21} \mbox{s}$$

The difference from the prior result (post #2) comes from using $=\frac{\hbar}{2}$ instead of $\approx\hbar$ which is cusomery as well.


Daniel.


------------------------------------------------------
[1] J.J.Sakurai,"Modern Quantum Mechanics",any of the 2 editions.

thankyou Daniel,

I see now. But why did you use 4pi instead of 2pi?

thanks,
Jason

 

[dextercioby]

It comes from

$$\frac{\hbar}{2}=\frac{h}{4\pi}$$.​

Daniel.