tys - nuclear physics and radioactivity

8/14/2019 TYS - Nuclear Physics and Radioactivity

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Nuclear Phvsics & RadioactivitvN93/10ic),(d)An approximate formula for the radius R of a nucleus of a nucleon number {mass number),4 is

R = (1.2 x 10 1s),4:where R i\ expressed in metres.(4 Use this lor muld lo calculate the radius fa )!Fe nucleus.(ii) Assuming the nucleus to be a uniform sphere, estimate the density of nuclear matter in

the iron nuLleus. Iakethe massol ::Fp nucleuslobe55.92O7r.(iii) Sketch a graph to show how the density p of nuclear matter depends on the nucleon

(c)

(c) (i) ror !!Fe, the mass nr-rmber A is s6. Hencen = (1.2 x 10-1s)(56)a = 4.591 x 10

1s - 4.6 x 1o-ts m(ii) The volume ofthe nucleus iassumed to be a un iform sphere), isv =!n7z = !n14.5g7 v1o-1s)3 = 4053 x 10 43 mr

number,4 of the nucleus.(d) The mass of 1 m3 of iron is B x 103 kg. There is a general trend for the densities of solidelements to increase with increasing nucleon number, Comment on these two pieces of information inrelation to our answers to (c)(ii) and (iii) and suggest an explanation for any difference between thebehavior ofthe density of solid elements and the density of nulear matter.

(iii)Density = Mass / Volume = = 2.290 x 1.017 = 2.3 x 1017 ksm 3lheformulafordensityrtp -i - . .L 2.3 l0r l.gm'r"l','" ,1where is the Lrnified mass constant (z = 7.66 x 10 2r kg).This shows that ihe density of nuclear matter is independent of mass number A.Graph of density p vs mass number,4 is:

p /x 1017 kF.n'l


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(d) Nuclear matter's density is many orders of magnitude greater than that of solid elements.Furthermore, whereas solid element's density increases with nucleon number, the density of nuclearmatter remains constant regard ess of nucleon number,The rason for this is that solid elements have a lot of empty space wlthin them. Rutherford's model ofthe atom (Geiger and Marsden's alpha scattering experimen0 showed that matter is mostly emptyspace, with most of the mass concentrated in the tinY nuclei. This accounts for the very low density ofmatter, such that the densest element known to man has a density of only 2 x 10a kgm r.This large amount of empty space ln so id elements means that there is much room for the nuclei t getcloser together, and thir density to increase tremendously, if somehow they can be compressed further(e.9. in an extremely sirong gravitational field of a black hoie).Within the nucleus, however, all the nucleons (i.e. protons and neutrons) are as closely packed as theycan be. With the proions and neutrons packed side-by-side within the nucleus, thre is no further spacefor them to compress. Therefore, nuclear matter of all elements has the same density of 2.3 x 1017

Ns6/s(c)Fig. 2 lists the atomic mass M and the binding energy B of some nuclides-

Nuclide B':N 15.0001 1 115.5r2o 1S 99492 't27.6'lo 17 99916 13 9.8,IF 18.99841

22.94777147.8

i 1rua 186.6?Xua 23.98505 198.3o,2e. 39.96238 343.B\2ca 39 96259 342.Otlca 43 95549 380.9\1sc 44.95592 387.8\'.ri 45.95263 398.2

(a) ln nuclear physics, nuclei with equal nucleon numbers but different protonnumbers are called bobors.ldenti{ytwo isobars in Fig.2.

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t1l(b) Although isobars must have the same nucleon numbers, their atomic masses can differ becauseof a difference in binding energy. confirm that, for the two isobars you have identified in (a), the heaviernucl,de of the pair has a smol/er binding energy than the lighter one. t1l


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(c) Would it be possible for the heavier nuclide of a pair of isobars to have a greater binding energythan the lighter one? Explain your answer.,ji(d) suppose that a,Zingle proton were removed from each

AO, tel, ilM g, )iS r. ^,1n . u,e tntor nat ion from I ig. 2 to calculate theWhat trend, if any, can you detect in the values ofthe energy?

I5lof the following nuclei:energy required in each case,

14t

Take: Speed of lightElementary chareerlLJnified atomic mass unitElectron massNeutron mass

c = 2.gg7g x lOB ms1e = 1.6022 x '\1-re cIL = 1.6605 x 10 'z7 kgm"=0000549zm' = 1.008665 amp = 1007276 u

(ai aoAr andaoca(b) The heavier nuclide is aocd, which has an atomic mass of 39.96259 z (heavieris 39.96238 r). aoca has a smaller binding energy of 342.0 MeV (smaller than aozlr

than aoAr wh ichwhich has 343.8

because bindingMeV).(c) Normally, the heavier nuclideenergy is also equal to the smountconstituent protons and neutrons,

is expected to have a smaller binding energyof energy required to completely separate a

fAlternative defin ition of binding energy:The amount of energy released when protons and neutrons combine to Jorm the nucleus.The lower the binding energy, the less energy released and hence the lower the mass defect (thedifference between the mass of the nucleus and its individual constituents). With Iess mass convertedinto energy, the nuclide is expected to retain more mass and hence be heavier.lHowever, it is still possible for a heavier nuclide to have a greater binding energy, if its constituents wereheavier to begin with.Neutrons are slightly heavier than protons. Hence if a nuclide has a larger neutron-to-proton ratio, itcould still end up heavier than its isobar with a lower neutron to proton ratio, even if its binding energyis higher.


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(d) 12.10 MeV,8.000 MeV,11.70 MeV,6.900 MeV,10.40 MeVThese energies, known as proton separation energies, are greatest when the ratio of proton : neutron is1 : 1 (see table below):

Nuclide No. ofNeutrons Proton SeparationEnergy / Mev,Fo 8 a 1.0 72_101luo 72 !2 10 al.ton1r i 22 24 0.92 10 401:F 9 10 0.90 8 000

Xlsc 21 74 0.88 6 900

The reason is that these nuclides in th table have low proton numbers ranging from 8 to 22. Nuclideswith low proton numbers are most stable when their proton : neutron ratio is 1 : 1- This is unlike heaviernuclides whereby the proton:neutron ration must be lessthan 1:1in orderto bestable.since 130 and trtMg hoth ha\e a proton : neutron ratio of 1 : 1, they are the most stable amongst thenuclides listed above. Their proton separation energy is therefore the greatest.

3 Ns2/s{b),(32c)(b) TheradionuclidelSPdecaysbytheemissionofasinglep particleto a stable dau8hternuclide.A physicists measures the activhy of a ]3P source by detecting and recording the number ofP particleemitted by the source in each of the five consecutive 10 - second intervals. The detection system givesthe following results:

Durat'ons/s Number off -particles

7 10 10212 10 9833 10 10034 10 9985 10 7047

(i) These results show no significant decrease with time. What can the physicists deduceaboutthe half life ofthe radionuclide? t1l(ii) The number of p particles emitted in each time interval fluctuates. Suggest an "..Mexplanation. (The error in the timing is negligible., ,t, u n ' ' '(iii) Deduce the mean activity ofthe source during this experiment. 121


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The physicist repeats the experiment 7-o0 days latet and finds that the mean ativitY of thesource has decreased by 28.8 %. A'A'l'rt.. \'1(iv) Find the hatftife of iip. 0 ?'2 '!a : A e tsl(c) A second source contains a mixture of the radionuclides llP and 1;P. The half-life of:3P is 1.77times longer than that of ]!P. et time t = 0, 90 % of the decays from this source come from the]!P component.(i) lftherearet{atomsof 1?Patthistime,howmanyatomsof iSParethere? t3l(:i) How long must elapse belore 90 % ofthe decays come from the 1?P ? l4l

(b) (i) Ihe half life is much longer than the duration of this experiment- The number ofradioactive nuclei remains approximately constant during this period, and hence the activity alsoremains approximately constant (since activity,4 = ,lN where 2 is the decay constant and N thenumber of radioactive nuclei remaining).(ii) Radioactivity is a random process. This means that it is not possible to predid withabsolute certainly which nucleus or whn a particular nucleus will decay in a certain timeinterval. This accounis for the fluctuation.(iiD Mean activity = 5052 + 50 - 101 s'(iv) lnitialactivity=100s1 After Tdays,theactivitybecameTl-2s1(28.8%reduction).

71.2 = looe ^'72 = 0.048525 day I4 = 0.0a8525

z

t. = 14.2a4 = 1.4.3 days

(d (i) Let the total activity be ,4.Usins,4 = 2N,For32P, o.gA = tryN ---- l7l


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For 33P. o.lA = tEN", --- 12\f!"(1) + (2) -:=#-1^ eN'9=-X.:_-Y-'ta2 N31 tt2\3=+=o1s7N'

lii) A32= AB2e-^=,t (3) whereAo'32 was its initialativity.A:B: ADFe-\"t (4) where Ao,33 was its initialativity.tsz = 14284daYs1," = -!a = 6.64s526,6,'i).-= t"2 = o.o27416dav'"' 771, t{{41+ {31 An -!}4e^r- ^Yr

o t.tootaszo-oottttolr79t = 208 days

4 N9t/a7ln an experiment carried out in 1903, Pierre Curie found that a sample of 4.1 g of pure radium {nowknown to be Radium 226) emitted energy at a rate of 0.12 W. Assuming that this energy is entirely inthe form of kinetic energy of the emitted d ' parricles, estimate the energy with which the particlesemerge- The half-life ofRadium-226 is 5.0 x 1010 s.

The decay constant,l = f =ffi=f.:eO"fO t'.'The number of radioa.tive nu(leiN - 0004I = 1.092 \ 10-'zActivityA =,lN = 1.386x 10-11x 1.092x 1022 = 1.514x 1011s1Since energy is emitted at a rate of 0.12 Js l the kinetic energy of each a - particle = --g


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= 7 .929 x \O-13 J= 7.9 x l0-r3 J

5 N01/QsTwo radioactive samples, A and B, have halflives of 5.0 hours and 10.0 hours respectively. At time t =0,Acontains 5.0 x 10 10 mol of undecayed nuclei, and B contains 1.0 x 10-10 mol.

(a) Calculate the ratio ofthe activity ofA to that of B at t = 0. t3l(b) At a certain time aftert = 0, the activities ofA and B become equal. Calculatethistime. [3]

(a) Activity ofA =,lN :43 x 5.0 x 1O-1o 'Nr\Activity ofA: ,1N : 4? x 1.0 x 10-10 { l'/tRat;o ofactivity ofAto that ofB at Activity ofA t = 0 is:

43 x S.0 x 10-10 : E? x 1.0 x 10 1010:1

(b) A.tivibr A: Aoe-^tLet the initial activity of B be r, then the initialactivity ofA will be 10a.Aftertime t, when both activities are equal,

nt"-#' = ,e toE'19" 'n', - "'l;'

Lnto -t!1t = -!!tr'z(i-n) = r, ro

t=33.219h:33h

tys - nuclear physics and radioactivity

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