The muon is a subatomic particle with the same charge as an electron but with a
ID: 1768452 • Letter: T
Question
The muon is a subatomic particle with the same charge as an electron but with a mass that is 207 times greater: mmu=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 mu s 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->1and 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.
mass of themuon mmu =207me
mass of the electron me =9.1*10-31kg
Planck's cosntant h=6.62*10-34Js
charge of the electron e =1.6*10-19 C
Rydberg constant R =10.97*106/m
Permittivity constant eo=8.9*10-12C2/N-m2
a) 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. (Got 2.27 * 10-24 m --> wrong)
b) What are the speed of a muon in the n=1 ground state?
c) What is the wavelength of the 2->1 muon transition?
d) Is the photon emitted in the 2->1 transition infrared, visible, ultraviolet, or x ray?
e) How many orbits will the muon complete during 1.5 mu s?
f) Is the answer for part E a sufficiently large number that the Bohr model "makes sense," even though the muon is not stable?
Please work the question out step by step
Explanation / Answer
(a)
r(muon) = a0 * n^2 / Z where a0 = bohr radius and n= 1
a0 = (h/2pi)^2 / (k*e^2*m)
So,
epsilon0 = permittivity constant
k = 1/4*pi*epsilon0 = 8.99 x 10^9 Nm^2/C^2
r(muon) = (h/2pi)^2 / (k*e^2*m(muon)*Z )
= (6.62 x 10^-34 / 2*3.14)^2 / (8.99 x 10^9 * (1.6 x 10^-19)^2 * 207*9.1094 x 10^-31 * 6)
= 4.26 x 10^-5 nm
(b) v = (Z/n) * a0 , n = 1
v = 6 * (6.62 x 10^-34 / 2*3.14)^2 / (8.99 x 10^9 * (1.6 x 10^-19)^2 * 207*9.1094 x 10^-31)
v = 1.31 x 10^7 m/s
(c ) En = -13.6 * Z^2 / n^2
Due to the Z^2 term, the energy
levels will by 6^2 times greater in magnitude compared to hydrogen. Also, ER = ke^2/2a0 and replacing the electron with the muon the Bohr radius gets smaller by a factor of 207.
Therefore, the magnitude of the energy levels will increase by a factor of 207 from this effect.
En = -13.6 * Z^2 /n^2 * 207
E1 = -13.6 * (6)^2 * 207 = - 0.101 MeV
E2 = 0.101 / 4 = - 0.0253 MeV
So,
E2 - E1 = -0.0253 + 0.101 = 0.0757 MeV
So,
lambda = hc/E = 6.62 x 10^-34 * 3 x 10^8 / 0.0757 x 10^6 = 0.0164 nm
(d) It is in X-ray region
(e)
No of orbits = v*t /(2*pi*r) = 1.31 x 10^7 * 1.5 x 10^-6 / (2*3.14 *4.26 x 10^-14)
N = 7.3 x 10^13 orbits