1000 Solved Problems in Modern Physics

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)