What is the problem with the Stern Gerlach J=1 exercise?

[folgorant]

Hello! i have a trouble with an exercise of a course of structure of matter.

Helium atoms in the excited state 2³S₁ come out from a box at v=2000 m/s and pass trough an Stern & Gerlach apparatus with dB/dz=25 T/m and length L=0.30m. Atoms spends 1 hour in the box and the average-time for decay is 8000s.
the questions are:

1) evaluate (number of atoms deflected with Mj=1) / (total number of atoms came out from the box)

2) evaluate the angular deflection of the component Mj=1.


So, the part 2 is very simple:
i know that the state has S=1, L=0, J=1 and using F = \mu_z * dB/dz I found alpha=0.30.

But I can't understand part 1:
8000s>1hours so every atom producted come out form the box and don't decay to the ground state??
then, don't the 3 components Mj= 1, 0, -1 should have the same probability 33.3% ?
Or I should use statistical mechanics and evaluate the probability of that state in function of temperature and energy?

please help me to understand, thank you!

 

[Redbelly98]

But I can't understand part 1:
8000s>1hours so every atom producted come out form the box and don't decay to the ground state??

This seems irrelevant, but with an exponential decay process some atoms will have decayed, some will take longer than 8000 s to decay. 8000 s is the average decay time.

then, don't the 3 components Mj= 1, 0, -1 should have the same probability 33.3% ?

Sounds right to me. So the answer is _____ ?

 

[folgorant]

the answer is: 21.25%

 

[Redbelly98]

Okay, my mistake. After reading the original post more carefully, we need to divide the J=1 atoms by the TOTAL number of atoms. So the decay part of the problem is relevant after all.

So, you first need to figure out what percentage of the atoms are remaining in the 2³S₁ state. It's an exponential decay type of problem.

 

[folgorant]

yes...
if N is total number of atoms;
y=1/8000 is the decay rate (number of decay/sec)
t= 1 hours = 3600sec is the period atoms be in the box

so: 1/8000 * 3600 * N = 0.45N i.e. the percentage of atoms remaining in excited state is 55%.

after this..i should evaluate the probably P(Mj=1)=[exp(-beta*E(Mj=1 state)]/[exp(-beta*E(Mj=1 state) + exp(-beta*E(Mj=0 state) + exp(-beta*E(Mj=-1 state)]...

where:
beta=1/(kT)
k= Boltzmann constant
T obtained from: 3/2 k T = 1/2 m v^2

but i should know the value of magnetic field to evaluate the energy states! instead i know only its gradient!

?

 

[Redbelly98]

yes...
if N is total number of atoms;
y=1/8000 is the decay rate (number of decay/sec)
t= 1 hours = 3600sec is the period atoms be in the box

so: 1/8000 * 3600 * N = 0.45N i.e. the percentage of atoms remaining in excited state is 55%.

No, the population of excited atoms is an exponentially decaying function. You are using a linear function, which is incorrect.

Using your formula, how many ground-state atoms are there after 8800 seconds? Look at the result you get ... do you see the problem?

after this..i should evaluate the probably P(Mj=1)=[exp(-beta*E(Mj=1 state)]/[exp(-beta*E(Mj=1 state) + exp(-beta*E(Mj=0 state) + exp(-beta*E(Mj=-1 state)]...

where:
beta=1/(kT)
k= Boltzmann constant
T obtained from: 3/2 k T = 1/2 m v^2

but i should know the value of magnetic field to evaluate the energy states! instead i know only its gradient!

?

Usually in this type of apparatus, B=0 immediately after leaving the oven. It is only after they have traveled some distance away from the oven exit that they enter the magnetic field.

 

[folgorant]

thank you very much Redbelly98...now it's okay!
i've used: N=N(0)exp(-t/T) and then dividing for 3 the result is correct!

thanks...

 

[Redbelly98]

You're welcome!