Ground State of Deutron | Physics Optional Notes for UPSC PDF Download

 Page 1


Ground state of Deuteron:
Deuteron is a system of two 
particles 
neutron
and 
proton 
of nearly equal.masses
le m1 
=m2=M
The reduced mass of system 
= 
m1m2 / 
(m1+m2) 
= 
M x
M/(M+M) 
=UtZ.
The force between two 
particles 
depends on distance
between two 
particles 
and acts along their line 
joining.
The 
potential 
energy can be expressed as V=V( r)
which is a square well potential 
and independent of
time,
The time independent schroedinger equation in
centre of mass system can be written as
v2w+ 2u(E-V(r) 
)tP=o 
---(1 
)
h2
where 
p is the reduced mass and E is total energy of
the system.
Equatign (1) can be expressed in'terrns of spherical
polar coordinates as
1a(fa-,)+ 1 D(sinod-.) 
+
p2 ap Dp plo,'',,2e ,ie ao
1 6:- 
+ 
Z--f-E-V( 
r)) 
-=g-12;
fsin2e ao h2
ln central force. the wave function 
g 
can be expressed
using spl"errca i harnionic finction
,iu -(U.-) 
v_ ($ 
c)
(
Page 2


Ground state of Deuteron:
Deuteron is a system of two 
particles 
neutron
and 
proton 
of nearly equal.masses
le m1 
=m2=M
The reduced mass of system 
= 
m1m2 / 
(m1+m2) 
= 
M x
M/(M+M) 
=UtZ.
The force between two 
particles 
depends on distance
between two 
particles 
and acts along their line 
joining.
The 
potential 
energy can be expressed as V=V( r)
which is a square well potential 
and independent of
time,
The time independent schroedinger equation in
centre of mass system can be written as
v2w+ 2u(E-V(r) 
)tP=o 
---(1 
)
h2
where 
p is the reduced mass and E is total energy of
the system.
Equatign (1) can be expressed in'terrns of spherical
polar coordinates as
1a(fa-,)+ 1 D(sinod-.) 
+
p2 ap Dp plo,'',,2e ,ie ao
1 6:- 
+ 
Z--f-E-V( 
r)) 
-=g-12;
fsin2e ao h2
ln central force. the wave function 
g 
can be expressed
using spl"errca i harnionic finction
,iu -(U.-) 
v_ ($ 
c)
(
second
Ld!_+2_sz tE-v(r)l 
=
U 
0r2 h2
-l 1 d{sin8iY,,-I€d)}+ r Aryn-I
Y'm (S,0) a0 de Yr,(6.d) sin2Q Q42 
--'121
ln the above equation LHS is function of r only and
RHS is a function of and only. Consider each
rjde 
,of 
the above equation is equal to i(l+1)
{ 
tV'- 
+ 
zuf, 
[E-V(r 11 = 
; 
1 
I 
.. 
11 
--(4)
Uaf 
h2
e.nd -.. t 
I 
A 
tsing 
AYi-_{e.dtr 
}* 
r dV
Yh, (g6I aa ag strfs a+2
=ltl+
l(l+ 1) ---(s)
Equations ( 4) and (5) can be written as
lUr
d 
{sin6 
3Yrm-jl€,r!L}+ 1
a0 
-AO_' 
sin,e
= 
0 
--(6)
I(l 
+t1Y,," 
16,a;
= 
0 ---(7)
I Y,. 
(e.o) 
)
a62
' V(r 
) 
(Ur_) 
Yr. 
(8,d)
r
/13
Page 3


Ground state of Deuteron:
Deuteron is a system of two 
particles 
neutron
and 
proton 
of nearly equal.masses
le m1 
=m2=M
The reduced mass of system 
= 
m1m2 / 
(m1+m2) 
= 
M x
M/(M+M) 
=UtZ.
The force between two 
particles 
depends on distance
between two 
particles 
and acts along their line 
joining.
The 
potential 
energy can be expressed as V=V( r)
which is a square well potential 
and independent of
time,
The time independent schroedinger equation in
centre of mass system can be written as
v2w+ 2u(E-V(r) 
)tP=o 
---(1 
)
h2
where 
p is the reduced mass and E is total energy of
the system.
Equatign (1) can be expressed in'terrns of spherical
polar coordinates as
1a(fa-,)+ 1 D(sinod-.) 
+
p2 ap Dp plo,'',,2e ,ie ao
1 6:- 
+ 
Z--f-E-V( 
r)) 
-=g-12;
fsin2e ao h2
ln central force. the wave function 
g 
can be expressed
using spl"errca i harnionic finction
,iu -(U.-) 
v_ ($ 
c)
(
second
Ld!_+2_sz tE-v(r)l 
=
U 
0r2 h2
-l 1 d{sin8iY,,-I€d)}+ r Aryn-I
Y'm (S,0) a0 de Yr,(6.d) sin2Q Q42 
--'121
ln the above equation LHS is function of r only and
RHS is a function of and only. Consider each
rjde 
,of 
the above equation is equal to i(l+1)
{ 
tV'- 
+ 
zuf, 
[E-V(r 11 = 
; 
1 
I 
.. 
11 
--(4)
Uaf 
h2
e.nd -.. t 
I 
A 
tsing 
AYi-_{e.dtr 
}* 
r dV
Yh, (g6I aa ag strfs a+2
=ltl+
l(l+ 1) ---(s)
Equations ( 4) and (5) can be written as
lUr
d 
{sin6 
3Yrm-jl€,r!L}+ 1
a0 
-AO_' 
sin,e
= 
0 
--(6)
I(l 
+t1Y,," 
16,a;
= 
0 ---(7)
I Y,. 
(e.o) 
)
a62
' V(r 
) 
(Ur_) 
Yr. 
(8,d)
r
/13
' 
The last term in equation (6) is called
oentrifugal 
potential 
because its derivative with ralspect
to r is equal to classical centrifugal force when angular
momentum is the centrifugal 
potential 
versus r
graphs for I 
= 
O,1,2 are shown in fi91. For binding the
proton and neutron together, the 
potential V (r) must be
attractive at least over a llmited range of r and should
have the value at least to compensate the repulsive
certifugal 
potential. This can be achieved in l=0 state.
The lowest 
quantum 
mechanical state for a fu/o body
system like deuteron is always an I 
= 
0 state.
L rilh:
.lmr'
The 
potential function 
providing the simplest
solution of schroedinger's equation is square well as
shown in fig.2 and expressed as V 
= 
-Vo for r<=ro 0
for r>ro
> 
r80
.':
a
{ 
roo
=
I
'i
(.,
Page 4


Ground state of Deuteron:
Deuteron is a system of two 
particles 
neutron
and 
proton 
of nearly equal.masses
le m1 
=m2=M
The reduced mass of system 
= 
m1m2 / 
(m1+m2) 
= 
M x
M/(M+M) 
=UtZ.
The force between two 
particles 
depends on distance
between two 
particles 
and acts along their line 
joining.
The 
potential 
energy can be expressed as V=V( r)
which is a square well potential 
and independent of
time,
The time independent schroedinger equation in
centre of mass system can be written as
v2w+ 2u(E-V(r) 
)tP=o 
---(1 
)
h2
where 
p is the reduced mass and E is total energy of
the system.
Equatign (1) can be expressed in'terrns of spherical
polar coordinates as
1a(fa-,)+ 1 D(sinod-.) 
+
p2 ap Dp plo,'',,2e ,ie ao
1 6:- 
+ 
Z--f-E-V( 
r)) 
-=g-12;
fsin2e ao h2
ln central force. the wave function 
g 
can be expressed
using spl"errca i harnionic finction
,iu -(U.-) 
v_ ($ 
c)
(
second
Ld!_+2_sz tE-v(r)l 
=
U 
0r2 h2
-l 1 d{sin8iY,,-I€d)}+ r Aryn-I
Y'm (S,0) a0 de Yr,(6.d) sin2Q Q42 
--'121
ln the above equation LHS is function of r only and
RHS is a function of and only. Consider each
rjde 
,of 
the above equation is equal to i(l+1)
{ 
tV'- 
+ 
zuf, 
[E-V(r 11 = 
; 
1 
I 
.. 
11 
--(4)
Uaf 
h2
e.nd -.. t 
I 
A 
tsing 
AYi-_{e.dtr 
}* 
r dV
Yh, (g6I aa ag strfs a+2
=ltl+
l(l+ 1) ---(s)
Equations ( 4) and (5) can be written as
lUr
d 
{sin6 
3Yrm-jl€,r!L}+ 1
a0 
-AO_' 
sin,e
= 
0 
--(6)
I(l 
+t1Y,," 
16,a;
= 
0 ---(7)
I Y,. 
(e.o) 
)
a62
' V(r 
) 
(Ur_) 
Yr. 
(8,d)
r
/13
' 
The last term in equation (6) is called
oentrifugal 
potential 
because its derivative with ralspect
to r is equal to classical centrifugal force when angular
momentum is the centrifugal 
potential 
versus r
graphs for I 
= 
O,1,2 are shown in fi91. For binding the
proton and neutron together, the 
potential V (r) must be
attractive at least over a llmited range of r and should
have the value at least to compensate the repulsive
certifugal 
potential. This can be achieved in l=0 state.
The lowest 
quantum 
mechanical state for a fu/o body
system like deuteron is always an I 
= 
0 state.
L rilh:
.lmr'
The 
potential function 
providing the simplest
solution of schroedinger's equation is square well as
shown in fig.2 and expressed as V 
= 
-Vo for r<=ro 0
for r>ro
> 
r80
.':
a
{ 
roo
=
I
'i
(.,
v(r 
)
The schroedinger equation for I and ll regions in ['= 0
state may be expressed as
dzu(r) 
= 
2u (Vo-Ee) u(r) 
=0, 
rsro 
--(8)
d? h2
and gl2u(r) 
-2u Esu(r) 
=0,r>ro 
---(9)
dl h2
E 
= 
- Ee 
= 
- 2.2?5MeV whichisthe binding energy of
the deuteron
Substituting''lZgo-EB)/ h 
= 
Kr 
--(a)
And 
.,1 
. 
zp Ee I h 
= 
Kz 
---(b)
r,quations (8) and (9) take the form
d2u(r) 
+ 
K,2U(r) 
=0 
r s ro 
--(10)
df
d2u( r) - Kr2 U( r) 
= 
r) r'o 
--(11)
dF
the general 
solutions of equations (10) and (11) may be
expressed as
U 
1(r) =Asin 
Kl r 
+ 
Bcos K. r ---(12)
Page 5


Ground state of Deuteron:
Deuteron is a system of two 
particles 
neutron
and 
proton 
of nearly equal.masses
le m1 
=m2=M
The reduced mass of system 
= 
m1m2 / 
(m1+m2) 
= 
M x
M/(M+M) 
=UtZ.
The force between two 
particles 
depends on distance
between two 
particles 
and acts along their line 
joining.
The 
potential 
energy can be expressed as V=V( r)
which is a square well potential 
and independent of
time,
The time independent schroedinger equation in
centre of mass system can be written as
v2w+ 2u(E-V(r) 
)tP=o 
---(1 
)
h2
where 
p is the reduced mass and E is total energy of
the system.
Equatign (1) can be expressed in'terrns of spherical
polar coordinates as
1a(fa-,)+ 1 D(sinod-.) 
+
p2 ap Dp plo,'',,2e ,ie ao
1 6:- 
+ 
Z--f-E-V( 
r)) 
-=g-12;
fsin2e ao h2
ln central force. the wave function 
g 
can be expressed
using spl"errca i harnionic finction
,iu -(U.-) 
v_ ($ 
c)
(
second
Ld!_+2_sz tE-v(r)l 
=
U 
0r2 h2
-l 1 d{sin8iY,,-I€d)}+ r Aryn-I
Y'm (S,0) a0 de Yr,(6.d) sin2Q Q42 
--'121
ln the above equation LHS is function of r only and
RHS is a function of and only. Consider each
rjde 
,of 
the above equation is equal to i(l+1)
{ 
tV'- 
+ 
zuf, 
[E-V(r 11 = 
; 
1 
I 
.. 
11 
--(4)
Uaf 
h2
e.nd -.. t 
I 
A 
tsing 
AYi-_{e.dtr 
}* 
r dV
Yh, (g6I aa ag strfs a+2
=ltl+
l(l+ 1) ---(s)
Equations ( 4) and (5) can be written as
lUr
d 
{sin6 
3Yrm-jl€,r!L}+ 1
a0 
-AO_' 
sin,e
= 
0 
--(6)
I(l 
+t1Y,," 
16,a;
= 
0 ---(7)
I Y,. 
(e.o) 
)
a62
' V(r 
) 
(Ur_) 
Yr. 
(8,d)
r
/13
' 
The last term in equation (6) is called
oentrifugal 
potential 
because its derivative with ralspect
to r is equal to classical centrifugal force when angular
momentum is the centrifugal 
potential 
versus r
graphs for I 
= 
O,1,2 are shown in fi91. For binding the
proton and neutron together, the 
potential V (r) must be
attractive at least over a llmited range of r and should
have the value at least to compensate the repulsive
certifugal 
potential. This can be achieved in l=0 state.
The lowest 
quantum 
mechanical state for a fu/o body
system like deuteron is always an I 
= 
0 state.
L rilh:
.lmr'
The 
potential function 
providing the simplest
solution of schroedinger's equation is square well as
shown in fig.2 and expressed as V 
= 
-Vo for r<=ro 0
for r>ro
> 
r80
.':
a
{ 
roo
=
I
'i
(.,
v(r 
)
The schroedinger equation for I and ll regions in ['= 0
state may be expressed as
dzu(r) 
= 
2u (Vo-Ee) u(r) 
=0, 
rsro 
--(8)
d? h2
and gl2u(r) 
-2u Esu(r) 
=0,r>ro 
---(9)
dl h2
E 
= 
- Ee 
= 
- 2.2?5MeV whichisthe binding energy of
the deuteron
Substituting''lZgo-EB)/ h 
= 
Kr 
--(a)
And 
.,1 
. 
zp Ee I h 
= 
Kz 
---(b)
r,quations (8) and (9) take the form
d2u(r) 
+ 
K,2U(r) 
=0 
r s ro 
--(10)
df
d2u( r) - Kr2 U( r) 
= 
r) r'o 
--(11)
dF
the general 
solutions of equations (10) and (11) may be
expressed as
U 
1(r) =Asin 
Kl r 
+ 
Bcos K. r ---(12)
u 
rr 
(r 
) = 
ce 
K2' 
* 
D e 
-K2' --(13)
The wave functions satisfied the condition that V(r 
) =
finite at r 
= 
0 and zero at r 
= 
o. The condition 9(r 
) =6
requires that U(r 
) = 
r. V vanishes at r 
= 
0 imirlies that
B=0.
Equation 
('12) 
takes the form
U1(r)=AsinKlr 
--('14)
The condition V(r) 
=g 
atr=o requiresthatc=0so
the U 
(r 
) 
remains finite.
Equation (1 3) takes the form
U 
rr 
(r 
) = 
D e 
-K2' 
-t('! 
5)
From the boundary condition, the first order derivative
with respect to r must be contihuous.
tPr= 
Vrr 0 Ur 
=Uri 
at r.s ro 
--(i)
dVr 
= 
dVu B 
dq =d!u 
atr=ro 
--(ii)
dr dr dr dr
condition 
(i) gives
A sin k1 ro 
= 
D e 
-K2 
to 
--(16)
A Kr cos kr ro 
=.Kz 
O e 
-K2'o -*(17)
Dlviding 
(17) by (16) we get
Kr cot K1 ro 
= 
-Kz --(18)
The above equation 
gives 
the relation between
binding energy Es of two nucleon system to range of
potential ro and the depth Vo.
Substitute the values of Kr and Kr in 
(18) 
, 
we 
get
CotKr ro 
= 
-KzlK1 -iEe/Vo -Ee
Ground state of deuteron:
The deuteron the only two nucleon syStem
and consists of one proton and one neutron. Deuteron
is used to explain some of the concepts involved in
discussing 
nuclear 
potential and 
quantum states of
nuclei. The 
prope(ies involved during the nuclear
interaction are