Energy width of ground state neutral pion

[joeybenn]

Homework Statement

Most of the particles known in physics are unstable. For example, the lifetime of the neutral pion, [tex]\pi^{0}[/tex], is about 1.0 x [tex]10^{-16}[/tex] s. Its mass is 1.35 x [tex]10^{8}[/tex] [tex]\frac{eV}{c^{2}}[/tex]. What is the energy width of the [tex]\pi^{0}[/tex] in its ground state.

Homework Equations

[tex]\Delta E[/tex][tex]\Delta t[/tex] = [tex]\frac{h_{bar}}{2}[/tex]

[tex]E_{n}[/tex]= [tex]\frac{n^{2}h^{2}}{8ml^{2}}[/tex]

The Attempt at a Solution

I was thinking of just putting the time into [tex]\Delta E[/tex][tex]\Delta t[/tex] = [tex]\frac{h_{bar}}{2}[/tex] and solving for [tex]\Delta E[/tex] but that seems too easy. Could I find the length and of the pion and use [tex]E_{n}[/tex]= [tex]\frac{n^{2}h^{2}}{8ml^{2}}[/tex] ?

I guess I am a little stumped at the moment.

BTW: I am not asking for an answer, just guidance.

 

[anathema]

Hello there,

You are on the right track with using the equations you have listed. To find the energy width of the \pi^{0} in its ground state, you can use the formula \Delta E\Delta t = \frac{h_{bar}}{2} as you mentioned. However, you will need to use the mass and lifetime of the \pi^{0} to calculate the energy width.

To start, you can rearrange the formula to solve for \Delta E. It should look like this: \Delta E = \frac{h_{bar}}{2\Delta t}. Then, you can plug in the values for h_{bar} and \Delta t. Keep in mind that the lifetime of the \pi^{0} is given in seconds, so you may need to convert it to a different unit (such as nanoseconds) to match the unit of h_{bar}.

Once you have calculated the energy width, you can compare it to the mass of the \pi^{0}. If the energy width is much smaller than the mass, then the \pi^{0} is considered to be in its ground state. If the energy width is similar to or larger than the mass, then the \pi^{0} is not in its ground state and may have undergone some decay.

I hope this helps guide you in the right direction. Good luck with your calculations!

 

[kerimek]

I would approach this problem by first understanding the concepts involved. The energy width of a particle is related to its lifetime and mass through the uncertainty principle. This means that the shorter the lifetime and the larger the mass, the larger the energy width will be.

In this case, we are given the lifetime and mass of the neutral pion, so we can use the uncertainty principle to calculate its energy width. The formula for the energy width is \Delta E = \frac{\hbar}{2\Delta t}, where \Delta t is the lifetime of the particle. Plugging in the values given, we get \Delta E = \frac{\hbar}{2(1.0 x 10^{-16} s)} = 5.27 x 10^{8} eV.

Another way to approach this problem is by using the energy levels of the pion. The energy levels of a particle are given by E_{n} = \frac{n^{2}h^{2}}{8mL^{2}}, where n is the principal quantum number, h is Planck's constant, m is the mass of the particle, and L is the length of the system. In this case, we can consider the pion as a spherical system with a radius of 1.35 x 10^{8} \frac{eV}{c^{2}} (since its mass is given in energy units). Plugging in the values, we get E_{1} = \frac{(1)^{2}h^{2}}{8(1.35 x 10^{8} \frac{eV}{c^{2}})^{2}} = 5.27 x 10^{8} eV. This is the same result we got using the uncertainty principle.

In conclusion, the energy width of the ground state neutral pion is 5.27 x 10^{8} eV. This means that the pion can have energies within this range and still be considered the same particle.