1000 Solved Problems in Modern Physics

(Grace) #1

3.3 Solutions 171


Fig. 3.7Deuteron wave
function and energy


whereμis the reduced mass=M/ 2 ,M,being neutron of proton mass. With
the assumption of spherical symmetry, the angular derivatives in the Laplacian
vanish and the radial equation is
1
r^2

d
dr

(

r^2

d
dr

)

ψ(r)+(M/^2 )[E−V(r)]ψ(r)=0(2)

With the change of variable

ψ(r)=

u(r)
r

(3)

Equation (2) becomes
d^2 u
dr^2

+

(

M

^2

)

[E−V(r)]u=0(4)

The total energy=−W, whereW=binding energy, is positive as the poten-
tial is positive
V 0 =−V, whereV 0 is positive
Equation (4) then becomes
d^2 u
dr^2

+

(

M

^2

)

(V 0 −W)u=0;r<R (5)

d^2 u
dr^2


MWu
^2

=0;r>R (6)

whereRis the range of nuclear forces, Fig. 3.7.
Calling
M(V 0 −W)
^2

=k^2 (7)

and
MW
^2

=γ^2 (8)

(5) and (6) become
d^2 u
dr^2

+k^2 u=0(9)

d^2 u
dr^2

−γ^2 u= 0 (10)
Free download pdf