Question on calculating mean range of alpha

[jackjoo87]

Li6(n,a)H3 , Q = 4.78 Mev. The H3 and He4 products are oppositely directed, away from the site of the compound nucleus, and hence they form one straight track.
How to calculate the mean range of the alpha particle if the atomic stopping power of helium realtive to air is given as 0.36?

 

[Bob S]

This paper gives a range-energy curve for alpha particles in air in Figure 1.

http://www.alpharubicon.com/basicnbc/article16radiological71.htm

Toward the end of the paper, this equation is given, which can be used for extrapolation:

Lapp and Andrews formula for Alpha range in air: (0.005 E + 0.285)E^3/2

A range-energy curve on page 650 of Evans "The Atomic Nucleus" gives the following ranges for low energy alphas in air:

1 MeV 0.5 cm
2 MeV 1.0 cm
3 MeV 1.6 cm
4 MeV 2.5 cm (This agrees with Fig. 1 in above URL)

In a cloud chamber, the alphas from alpha sources create easily visible tracks with very little range straggling.

Very roughly at higher energies, the stopping power dE/dx (Bethe-Bloch equation) of a particle of charge z losing energy in a medium of charge Z scales as z²Z.

Bob S

 

[Bob S]

1) The problem you are referring to is on page 667 in Evans, and is for 2 atm air; the answers I gave for alpha particle range in my previous post are for 1 atm air.

2) Evans wants you to use Equations 3.9 through 3.12 on pages 650-651. All the particles in the problem are non-relativistic, so use E = ½Mv². The alpha particle and the H³ have equal and opposite momentum, but unequal energies. Turn the crank.

Bob S

 

[jackjoo87]

thanks for your reply, Bob S.
I already calculate the answers for a(1.33) ,b (0.165), and c, KE(alpha)=2.0541 MeV); KE(H3)= 2.7259 MeV.
For question d, how mean length of alpha particle changes with pressure within the chamber? can i get directly from the curve Fig. 3.2?

 

[Bob S]

thanks for your reply, Bob S.
I already calculate the answers for a(1.33) ,b (0.165), and c, KE(alpha)=2.0541 MeV); KE(H3)= 2.7259 MeV.
For question d, how mean length of alpha particle changes with pressure within the chamber? can i get directly from the curve Fig. 3.2?

This is a question on page 667 of Evans "The Atomic Nucleus". The curve 3.2 for alpha particle range on page 650 is for 1 atm air. The question asks what the range would be in 2 atm helium. the Bethe-Bloch dE/dx equation (Evans page 637) is directly proportional to the product of the number N of atoms/cm³ and the number Z electrons per atom in the stopping medium. The range is inversely proportional to NZ (see Evans page 647). For a gas (ideal gas law), the number of molecules per cm³ is the same for all gases at 1 atm. Air (a diatomic molecule) has twice the number of atoms/cm³ at 1 atm as helium. Furthermore, helium has a Z of 2, while air has an effective Z of ~7.2 per atom. Does this help?

An alternative (and simpler) way of thinking about this is the Bethe-Bloch energy loss dE/dx due entirely to collisions with electrons, and is therefore proportional to the number of electrons per cm³. Using this guidance, NZ for air at 1 atm is proportional to 1 x 2 x 7.2, while for 2 atm helium it is 2 x 1 x 2. So the range in 2 atm helium is 14.4/4 = 3.6 times the range in 1 atm air. Does this make sense?

Bob S

 

[jackjoo87]

thank you very much for your help Bob S... :)
am i suppose to use equation from Pg 652 (eq 3.14)? because the book give us the relative atomic stopping power ratio...

 

[jackjoo87]

dr(H3)=N(air)/N(He) dr(air)/Sa
2x N(air) at 1 atm=N (He) at 2 atm. thus, dr(H3)=1/2 dr(air)/0.36. dr(air) i get from the curve.. am i correct?