The ortho-para conversion of deuterium and the electric quadrupole moment of the deuteron

The ortho-para conversion of deuterium and the electric quadrupole moment of the deuteron

Physica VII, no. 3 Maart 1940 THE ORTHO-PARA CONVERSION OF DEUTERIUM AND THE ELECTRIC QUADRUPOLE MOMENT OF THE DEUTERON by H. B. G. CASIMIR Commu...

383KB Sizes 0 Downloads 24 Views

Physica

VII,

no. 3

Maart

1940

THE ORTHO-PARA CONVERSION OF DEUTERIUM AND THE ELECTRIC QUADRUPOLE MOMENT OF THE DEUTERON by H. B. G. CASIMIR Communication

Suppl.

No. 89 from

the Kamerlingh

Onnes

Laboratory,

Leiden

Summary The influence of the interaction between inhomogeneous electric fields and the quadrupole moment of the deuteron on the ortho-para conversion of deuterium is discussed and it is shown, that the experimental results of F a r k a s and S a n d 1 e r on the conversion of Hz and D, in diamagnetic liquids can be explained by assuming the existence of a quadrupole moment of the order of magnitude found by R a b i .and his collaborators.

1. It is well known, that the conversion of parahydrogen into orthohydrogen can be catalysed by paramagnetic molecules or ions. The strongly inhomogeneous fields due to these ions will act differently on the two hydrogen nuclei and hence cause transitions between para- and ortho-states. In a paper on the kinetics of this reaction K al c k ar and Te lle rl) pointed out, that in the case of deuterium the nucleus might possess an electric quadrupole moment besides a magnetic moment and if this is the case also inhomogeneous electric fields would give rise to ortho-para transitions. At that time, however, there existed no experimental evidence for the existence of nuclear quadrupole moments, neither for the deuteron nor for heavy nuclei. To day we know the quadrupole moments of a number ‘of complicated nuclei with a fair degree of accuracy and very recently R a b i 3 and his collaborators, using the molecular beam resonance method have been able to ascertain the existence of a small but non vanishing quadrupole moment for the deuteron. The question arises, whether this moment will make itself felt in the ortho-para conversion. Now shortly before the first publication of Ra b i’s results, Far kas and San dle r “) have reported a number of experiments on the conversion of hydrogen Phyaica

VII

169 II’

If0

H.

B. G. CASIMIR

and deuterium dissolved in diamagnetic liquids, from which they conclude, that in the case of deuterium the conversion is not exclusively due to magnetic interactions, but that there must exist still another mechanism. Following the suggestion of K a 1 c k a r and T e 11 e r they assume, that this mechanism is connected with the.existence of a nuclear quadrupole moment. It is the object of the present paper to study this question in more detail and to investigate whether it is possible to account for F a r k a s and S a n d 1 e r’s results assuming a quadrupole moment of the order of magnitude found by R a b i C.S. 2. A complete calculation of the ortho-para conversion would be rather complicated. We will show that it is relatively simple to obtain an expression for the ratio of the conversion caused by a magnetic dipole and the conversion caused by an electric charge, if we assume that the orbital wavefunctions for the colliding particles are the same in both cases. The method used is analogous to that used by K a 1 c k a r and T e 11 e r r) for determining the ratio of the magnetic conversion for hydrogen and deuterium. We will first write down the interaction operators. Let

be the components of the magnetic moment of the colliding particle (either molecule, atom or ion), where J; are the components of the angular momentum (in units 5) and let xi, x2, x3 be the coordinates of a deuteron. The magnetic potential in a point with coordinates xi will be given by P=-Qmc

*Iit

i Jiai 0+

where ai denotes differentiation with respect to Xi and where Y is the distance between the magnetic dipole and the deuteron. The magnetic field will be given by:

and, if we write the nuclear magnetic &YSi

moment

in the form

THE

ORTHO-PAR4

CONVERSION

OF

171

DEUTERIUM

.

where M, is the mass of the proton and the si are the angular momentum matrices for the deuteron, then the total interaction operator will be given b.y

where the indices 1 and 2 refer to the first and the second deuteron in the deuterium molecule. This operator can also be written as

and only the first term will give rise to ortho-para transitions. will write this part ,of the interaction operator in the form. WA = C, fk Aik Ji [sf’-

We

$1

where

does not depend on the spins, and where the constant bY c,=-*gy&&.

C, is given

P

We now proceed to write down the operator of the interaction between an electric charge and a quadrupole moment. The energy -of an electric quadrupole in a field with potential cp(xJ is given by iFk ai%? fff

xixkpd7

where p is the charge density in the nucleus. This expression can be replaced by 4 Z &a,cp . sisk . --- Qc (22-l)

i,k

i

where

and i is the nuclear spin (for the deuteron i = 1). If the field is due to a charge ‘yt,e the interaction operator is given by V=*n,e2

(2i--’

~[(aiak~j14W

+

(aiakt),~P)si~)j.

172

H.

The operator in the form

giving

B. G. CASIMIR

ortho-para

transitions

VA = c, 5 A;k [s’i’) sp -

can be written

sp sp-j

with C, = Q n# e2 Q/(2i -

I)i.

In order to calculate the transition probability in the magnetic case we must take a matrix element characterized by the quantumnumbers m, m’ of the magnetic ion or molecule, the quantumnumbers a, b referring to the orbital motion and the nuclear spin quantum numbers. The general matrix element will be of the form Cgz @Wilm’)

(WLkI4

WY)--sk2)IP).

We must take the square of this matrix element, sum over all m’ and take the average over all initial states m, sum over all final parastates, take the average over all initial orthostates and finally, after multiplication with appropriate statistical factors, sum over a and b. In the electric case the corresponding matrixelement will be c, 2 (a 1Ajk 1b) (0 1s’i” sp - spsp ( p), We will now calculate the sums: p, = c2g&

z 2,: F ( 5 (a I AikI 4 (m I Ji I m’) (0I 4’-

4”’ I P) I2

and p, = cg : f 12 (621A;/$[ h) (0 ) sp sp -

s\2) SP’ 1p) (2.

Since we are only interested in the ratio of the transition probabilities we have summed over all o-states instead of taking the average. Further, since the (olo)and (p(p) matrix elements of s~‘)-sS~“) are zero, we may just as well sum over all initial and final nuclear spin states: this will only lead to a factor 2. So if v and v’ are indices referring to all nuclear spin states:

where Sp denotes the diagonal

sum. Now

SP (JiJd = 31’ (i -I- 1) Pi + 1h and

THE

ORTHO-PARA

CONVERSION.

OF

173

DEUTERIUM

. SPMk’)-

@I) (sg) -

$‘)I

= 4 i (i + 1) (2i + I)2 8k”,

and therefore 2p, -

C$% j (j + 1) 1: (i + 1) (2i + l)2i? 1 (UIAik(b)

(2.

In the electric case we find: 2p,=CZ&++4ik~

The products since

b) (a lA~mlb)*sp[(s~‘)sp)-s~2)sp’)

($‘s~‘-sps~‘)].

s(i’)sil) . sj2)sz) will lead to terms with aik . a[,,, and, X 8ik AIR = F Aii = 0 i,k

these terms do not contribute to p,. So we can also write 2p6 = 2C:; zn (ajA,lb) (aJA1,,,Ib)* Sp (s(i’)sk” s\‘) s$j I , where Sp is still the diagonal sum over all quantum states of two nuclear spins. If we denote by .Sp(‘) the diagonal sum over the quantum states of one spin we have Sfi (S(i')Sr)Sl')S!p!)) = (2i + 1) Sp”’ (sisksls,“)e We will use the following equations: Spcl) St =: SP(') s$ = sp'l'stj = + 212+ i 21

1 1 - --‘u---v SP’W~ = s P(Us23 I 3 -- s P(Us329 2 3 - 15 3. Sfic”S1S2S*S2

=‘Sp(l)S*S3SIS3

=

SPc”S2S3S2S3

=

1

-V2

15

+--V

2 15

with 21= i (i + 1) (2i + 1);

the diagonal sum of all combinations one of the si is zero. We find: 2$, = 2(2i f

1) cz:

( (UlAikIb)

For the ratio of the transition magnetic case we find

w, -= W, Substituting

j(j +

containing

an odd power of

!‘(&ZJ2+kzI).

probabilities

in the electric and

1j2

11g42(~)2(~)2.~j2(2i-

i e4n:Q2[ii(i numerical

+ 11 (2i + 1) +A]

values and taking

i = 1 we obtain:

_%U _ i(i + lk2u2 x o 020 -7” * W,

where Q is measured in 1O-24 cm2.



(1)

174

H.

I$. G. CASIIUR.

-

:

3. If it were possible to find a case where eq. (1) is. directly applicable, this equation should enable us to obtain accurate values of the quadrupole moment Q. Unfortunately such an application seems to be out of the question. In the gas-phase it will be impossible to obtain a sufficient concentration of ions and in solutions the effect of the electric fields of dissolved ions will always be smallcompared with the effect of the‘water dipoles. More promising is the case of conversion by an electric dipole gas and although eq. (1) is not rigorously valid in this case, we will show that it is sufficient for obtaining an estimate of the order of magnitude to be expected. We will first substitute the values of ,y and Q 284). y = 0.85, Eq. (1) reduces to

where 7,~.and T,,,~,~.are the half-life

Q = 2.5 x 10-3.

in the electric and in the magnetic a substance has an electric dipole moment p, the field a will be of the same order of magnitude as the field (p/a). Since most transitions will be caused by close (2) will give the correct order of magnitude if we write p = n,ea where P is the radius of the molecule. For CO for instance, n, will roughly be l/50. According to F a r k a s “) the half-life of parahydrogen in oxygen at atmospheric pressure is 1.9 minutes. The halflife of ortho deuterium is roughly ten times .as long, so that T,,,~~,,.- 20 min. For oxygen i(i + l)g2 = 8 and hence 1 ‘id. = 8 x 2300 x 20 x -109rnin. n3 for orthodeuterium in CO at atmospheric pressure. Even at a CO pressure of 1000 atmospheres the half-life is of the order of one year. If we take a substance like HCl instead of CO the ,,effective charge” n, will be about ten times as large but it would be impossible to use high densities. We arrive at the conclusion that it is impossible to observe the influence of the electric quadrupole moment on the ortho-para transition in the gas phase. It is of course a somewhat bold extrapolation to use eq. (2) for a comparison of the magnetic conversion in O2 and the electric conversion of deuterium dissolved in water. If we assume that the number case. Now if at a distance of a charge collisions eq.

THE’

ORTHO-PARA

CONVERSION

175 .

OF DEIJ’TERIUM

of collisions is proportional to the number of molecules per cc we find 8 x 2300 1 -X18 20 =zgmin Te1. = x 22.2 x 1000 72: * fiz In order to account for F a r k a s and S a n d 1 e r’s results Tag.should be of the order of 600 min. and therefore ne should be 0.7. The electric action of a water molecule is thus found to be equivalent to the action of a charge 0.7 e placed at the centre of an oxygen molecule. As far as order of magnitude is concerned this is a very plausible result. We will now try to arrive at a somewhat closer estimate of the electric conversion in water. In table I we have reproduced some of F a r k a s and S a n d 1 e r’s data. The conversion of parahydrogen TABLE

I

Hydrogen ,

Sol\wlt

5 min

Deuterium k

lit min-1 6.8 2.9 17.0 2.3 2.7

x x x x x

mol-1

: min

10-s 10-S 10-s 10-S IO-5

480 1500 a000 13000

)

it lit min-1 2.6 x 4.1 x 0.5 x 0.5 x

mol-1 10-s 10-s 10-s 10-s

in water is probably partly due to the magnetic field of the proton spins, partly to the field of rotational magnetic moments - which often happen to be of the same order of magnitude as nuclear moments - and perhaps also to some form of induced magnetism. If we assume, that the difference of the reaction constants in water and in heavy water is due to the difference of nuclear spin then it follows, that the proton spins lead to a reaction constant 4 x I Od5. The corresponding reaction constant for deuterium would be 0.4 x 10m5. Now the electric dipole moment of the watermolecule can be accounted for by saying that +j of the protoncharge is screened off. Therefore we will apply eq. (2) with 12, = 4, i = i and g 7 2 x 2.785/1840.. This leads to But in this- way we have only taken into account the existence of positive charges in the water molecule. The negative charges will exert an influence of the same order of magnitude, therefore

176

THE

ORTHO-PARA

CONVERSION

Tel.

=

OF

DEUTERIUM

%“lq”./4

and kel, = 1.6 x 1O-5. The magnetic conversion for deuterium will be 10 times slower than for hydrogen and be given by k = 0.7 x 10W5 so that we find for the total reaction constant: k = 2.3 x 1O-5

in surprisingly good agreement with the experimental value. Of course this agreement may be partly fortuitous, but there can hardly be any doubt, that the explanation of F a r k a s and S a n d 1 e r is correct and that the conversion of deuterium in water is mainly due to the action of inhomogeneous electric fields on the nuclear quadrupole moment of the deuteron. It would be desirable also to measure the conversion of deuterium in heavy water: the difference of the reaction constants in ordinary water and in heavy water should only amount to 0.4 x 10M5. The existence of an electric quadrupole moment of the deuteron has very fundamental consequences for the theory of nuclear forces. Therefore it is a result of some importance that the existence of such a quadrupole moment has now been verified by an independent method.

REFERENCES 1) F. I< 2) J. M. Phys. 3) L. 1: 4) .J. M. Phys. 5) A. F

n I c li n r and E. T e 1 I e r, Proc. roy. SW. 130, 520, 1935. B. Ii e I I o g. I. I. R n b i, N. F. R a m se y jr. and J. R. Z a c h R r i n s, Rev. T,B, 318, 1939. a r k a s and L. S a n d 1 c r, Trans. Far. Sot. 35, 337, 1939. B. I< e 11 o g, I. I. R a b i, N. F. R a m se y jr. and J. R. Z a c h n r i as, Rev. 56, 729, 1939. a r k a s, Ortho-, Para- and Heavy Hydrogen. Cambridge 1935.