Deuteron Transcendental Equation

[James_1978]

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.

 

[anuttarasammyak]

How much value of b you get ? Please let me know it for checking your result.

 

[James_1978]

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.

 

[anuttarasammyak]

b has no physical dimension.

 

[James_1978]

Yes. I saw that. b*R is unit-less. My mistake.

 

[anuttarasammyak]

Is MeV translated to MKSA Joule ?

 

[James_1978]

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?

 

[anuttarasammyak]

And you say R is 2 fm.

 

[James_1978]

Yes, I use 2x10-15 m.

 

[anuttarasammyak]

Or with R I get

So you say b=0.4856.

 

[James_1978]

Yes. That is what I got.

 

[anuttarasammyak]

The first crossing point is approximately x=3.75

By numerical solution x=3.7688.. How much V_0 does it give us?

[EDIT] I could read the text in https://www.chegg.com/homework-help...r-physics-3-condition-existence-bou-q51837337 . The author himself investigated V_0=35 Mev. 36 and 35 are not so much different for me.

 

[vela]

So you say b=0.4856.

Yes. That is what I got.

I got a different value for $b$. I calculated $b=0.4627$.

 

[James_1978]

What did you use for E and mass of proton?

 

[vela]

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 $R=2~\rm fm$.