Deuteron Transcendental Equation

In summary: I used the number you gave for the reduced mass of the proton and neutron- 1.673x10^{-27}~\rm kg- and got the two different values for b.
  • #1
James_1978
37
3
Homework Statement
I believe I solved the transcendental equation but the plot does not make sense.
Relevant Equations
##k_{1} \cot{k_{1}R} = -k_{2}##
##k_{1} = \frac{\sqrt{2m(E+V_{o})}}{\hbar}##
##k_{2} = \frac{\sqrt{-2mE}}{\hbar}##
##x = -\tan{bx}##
##x = \sqrt{\frac{-(V_{o} + E)}{E}}##
Dear Forum,

I am trying to solve a problem (4.6) from the introductory nuclear physics textbook by Krane. The problem is as follows:
Solving the deuteron using the radial equations gives the transcendental function,

##k_{1} \cot{k_{1}R} = -k_{2}##

Were

##k_{1} = \frac{\sqrt{2m(E+V_{o})}}{\hbar}##

And

##k_{2} = \frac{\sqrt{-2mE}}{\hbar}##

That gives the relations between and R. Show that this equation can be written in the form,

##x = -\tan{bx}##

Where

##x = \sqrt{\frac{-(V_{o} + E)}{E}}##

Evaluate the parameter b for R = 2fm. Note that is the reduced mass. Solve the transcendental equation.

When rearranging we get ##b## as.

##b = \frac{\sqrt{-2mE}}{\hbar}*R##

For the reduced mass ##m = \frac{1.67x10^{-27}}{2} kg##
For ##\hbar = 1.054x10^{-34} J-s##
For ##E = -2.22 MeV##

We are suppose to see that when solving the transcendental equation we get ##V_{o} = 36 MeV##. However we must have something wrong because the function does not clearly show how you infer the ##V_{o} = 36 MeV##. Any help is appreciated.
 
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  • #2
How much value of b you get ? Please let me know it for checking your result.
 
Last edited:
  • #3
Dear Anuttarasmmyak

Here is what I got for b

##b = \frac{\sqrt{-2*\frac{1.67x10^{-27}}{2}*-2.22*1.602x10^{-13}}}{1.054x10^{-34}}##

Where ##b =2.3155x10^{14} m^{-1}##

Or with R I get ##b*R = 0.4856## I get ##V_{o} = 36##. I think this is correct. Just wanted to make sure.

1676640075854.png
 
  • #4
b has no physical dimension.
 
  • #5
Yes. I saw that. b*R is unit-less. My mistake.
 
  • #6
Is MeV translated to MKSA Joule ?
 
  • #7
Yes. I think you are asking in that I multiplied MeV*1.602x10^-13 to convert MeV to Joules. Is that what you are asking?
 
  • #8
And you say R is 2 fm.
 
  • #9
Yes, I use 2x10-15 m.
 
  • #10
James_1978 said:
Or with R I get
So you say b=0.4856.
 
  • #11
Yes. That is what I got.
 
  • #13
anuttarasammyak said:
So you say b=0.4856.
James_1978 said:
Yes. That is what I got.
I got a different value for ##b##. I calculated ##b=0.4627##.
 
  • #14
What did you use for E and mass of proton?
 
  • #15
I first calculated ##b## using the numbers you used and got a different answer. So I looked up the mass of a proton and neutron and found the reduced mass (##1.673\times 10^{-27}~\rm kg##) and used that to get the number above. Either way, I didn't get the value for ##b## you found. In both cases I used ##E_1 = -2.22~\rm MeV##.

In fact, I don't get the same values for the calculations you showed in post #3 for ##b## (really ##b/R##) or ##b R## (really ##b##). Moreover, your two values don't make sense if you're using ##R=2~\rm fm##.
 

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