What is the orbital radius of a muon captured in the n=1 ground state of Carbon?

[eedftt]

Homework Statement

Mastering Physics

Relevant Equations

Quantization

The muon is a subatomic particle with the same charge as an electron but with a mass that is 207 times greater: mμ=207me. Physicists think of muons as "heavy electrons." However, the muon is not a stable particle; it decays with a half-life of 1.5 μs into an electron plus two neutrinos. Muons from cosmic rays are sometimes "captured" by the nuclei of the atoms in a solid. A captured muon orbits this nucleus, like an electron, until it decays. Because the muon is often captured into an excited orbit (n>1), its presence can be detected by observing the photons emitted in transitions such as 2→1 and 3→1.
Consider a muon captured by a carbon nucleus (Z=6). Because of its large mass, the muon orbits well inside the electron cloud and is not affected by the electrons. Thus the muon "sees" the full nuclear charge Ze and acts like the electron in a hydrogen-like ion.

1）What is the orbital radius of a muon in the n=1
ground state? Note that the mass of a muon differs from the mass of an electron.
Express your answer with the appropriate units.My solution is:

The potential energy is: −Ze^2/4π*ϵ0*r and -13.60*Z^2* eV/n2 (n=1)

so r = e^2/(4π*ϵ0*13.60*Z*e) =1.76*10^(-11)

I do not know where I get wrong. Thanks!

 

[etotheipi]

A few things to clear up first; the relation $E_n = \frac{-13.60Z^2}{n^2}$ applies to hydrogen-like species with electrons, not muons. And it is also worth noting that $E_n$ is the total energy of the orbit (kinetic + potential), and not just the potential energy.

It might be a good idea to try and derive the relevant expressions from scratch. All that you need to know to get going is that that angular momentum is a multiple of $\hbar$, i.e. $L = mvr = n\hbar$.

Your potential energy expression is correct. You might be interested in the virial theorem.

 

[eedftt]

A few things to clear up first; the relation $E_n = \frac{-13.60Z^2}{n^2}$ applies to hydrogen-like species with electrons, not muons. And it is also worth noting that $E_n$ is the total energy of the orbit (kinetic + potential), and not just the potential energy.

It might be a good idea to try and derive the relevant expressions from scratch. All that you need to know to get going is that that angular momentum is a multiple of $\hbar$, i.e. $L = mvr = n\hbar$.

Your potential energy expression is correct. You might be interested in the virial theorem.

But in this question v and r are both unknown value. Thanks!

 

[etotheipi]

Well yes, the $r(n=1)$ is what you're trying to find.

How about this: can you write down the kinetic energy of the muon? It helps to know that $E_k = \frac{-U}{2}$ if $U$ is the potential energy. Can you equate this to another expression for kinetic energy?

Then, how might you go about eliminating $v$?

 

[eedftt]

Well yes, the $r(n=1)$ is what you're trying to find.

How about this: can you write down the kinetic energy of the muon? It helps to know that $E_k = \frac{-U}{2}$ if $U$ is the potential energy. Can you equate this to another expression for kinetic energy?

Then, how might you go about eliminating $v$?

So 1/2*m*v^2=13.6*Z^2/2

and v=(13.6*Z^2/m)^(1/2)
= 1.6*10^15 m/s

that is even more than speed of light.

 

[etotheipi]

No, forget the $\frac{-13.60Z^2}{n^2}$.

One expression for kinetic energy is, as you said, $\frac{1}{2}mv^2$. Can you use the virial theorem to find another expression for kinetic energy - you already have an expression for the potential energy? Alternatively, you could apply some circular motion expressions, but this is a slightly longer route.

Once you have those two expressions, you can eliminate $v$ using the angular momentum condition and solve for $r$.

 

[PeroK]

Note that the mass of a muon differs from the mass of an electron.

How did you use the different mass of the muon?

 

[eedftt]

No, forget the $\frac{-13.60Z^2}{n^2}$.

One expression for kinetic energy is, as you said, $\frac{1}{2}mv^2$. Can you use the virial theorem to find another expression for kinetic energy - you already have an expression for the potential energy? Alternatively, you could apply some circular motion expressions, but this is a slightly longer route.

Once you have those two expressions, you can eliminate $v$ using the angular momentum condition and solve for $r$.

My instructor does not tells us about virial theorem. Thank you!

 

[eedftt]

How did you use the different mass of the muon?

In this question the mass is 207me.

 

[PeroK]

In this question the mass is 207me.

Yes, I saw that. What was your calculation based on? What was the relevance of "potential energy"?

 

[kuruman]

Homework Statement:: Mastering Physics
Relevant Equations:: Quantization

so r = e^2/(4π*ϵ0*13.60*Z*e) =1.76*10^(-11)

I do not know where I get wrong.

You get wrong because you substituted 13.6 for the energy which is a number expressed in eV units while the rest of the numbers are (probably) in SI units. You might, I stress might, get the correct answer by converting 13.6 eV to the appropriate units. I recommend that you either derive or look up the expression for the Bohr radius in a hydrogen atom and adapt it to this case.

 

[etotheipi]

You get wrong because you substituted 13.6 for the energy which is a number expressed in eV units while the rest of the numbers are (probably) in SI units. You might, I stress might, get the correct answer by converting 13.6 eV to the appropriate units. I recommend that you either derive or look up the expression for the Bohr radius in a hydrogen atom and adapt it to this case.

But isn't the number $-13.60$ calculated from $m_e$, and not $m_{\mu}$? Granted, it's fairly easy to scale the formula directly provided you know how $E_n$ varies with mass in the algebraic formula.

 

[kuruman]

But isn't the number $-13.60$ calculated from $m_e$, and not $m_{\mu}$? Granted, it's fairly easy to scale the formula directly provided you know how $E_n$ varies with mass in the algebraic formula.

Yes, it is calculated from $m_e$ and one can scale the radius by placing the $207$ where it belongs in the expression for the Bohr radius. The roundabout way of using $13.6$ as a number instead of $E_1$ without dividing by the electron charge is likely to give an incorrect numerical result when combined with other quantities in SI units.

 

[eedftt]

You get wrong because you substituted 13.6 for the energy which is a number expressed in eV units while the rest of the numbers are (probably) in SI units. You might, I stress might, get the correct answer by converting 13.6 eV to the appropriate units. I recommend that you either derive or look up the expression for the Bohr radius in a hydrogen atom and adapt it to this case.

I actually consider the e. I think I just get the logic error. And from now on, I did not find my logic error.