PHYS420 (Spring 2002) Riq Parra Homework # 8 Solutions

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1 PHYS4 (Sprig ) Riq Parra Homework # 8 Solutios Problems. The ame uclide is a geeral term for ay isotope for ay elemet. Usig a table of uclides, fid all the possible aturally occurrig isotopes of xeo (Xe) ad cesium (Cs) alog with their relative abudaces. A olie table of uclides ca be foud i the class website uder Resources. For xeo we have ie differet aturally occurrig isotopes. Xe-4. % Xe-6.9 % Xe-8.9 % Xe % Xe-3 4. % Xe-3. % Xe % Xe-34.4 % Xe % For cesium we oly have oe. Cs-33 %. Startig with the expoetial law of radioactive decay, N( N e where λ is the decay costat, ad usig the defiitio of half-life ( T ), show that: l T / λ The half-life is defied as the time it takes half of a give umber of radioactive N uclei to decay. Mathematically, this coditio is defied as N ( T/ ). Pluggig this is for N( ad solvig for T / yields the desired expressio. N N e λt / l( ) λt l T / λ / 3. The decay rate R, defied as the umber of decays per uit time i a give dn radioactive sample, is give by R. The decay rate of a sample is ofte dt

2 referred to as its activity ad frequetly uses uits of curies (Ci), defied as Ci 3.7x decays / s. A sample of the isotope I 3, which has a half-life of 8.4 days, has a activity of 5 mci at the time of shipmet. Upo receipt of the I 3 i a medical laboratory, its activity is 4. mci. How much time has elapsed betwee the two measuremets? We calculate the decay rate usig the defiitio, dn( R( N λ e Re dt Solvig for t, R T/ R t l l λ R( l R( Pluggig i the umbers, 8.4 days 5mCi t l. days have elapsed. l 4. mci 4. The Actiium Series begis with the isotope 9 U 35 ; each atom seds out i successio the followig particles: α, β, α, β, α, α, α, α, β, β, α. From this iformatio ad by cosultig the periodic table, write out a accout of the Actiium Series (i.e. 9U 35 α? β? ) complete with the ecessary symbols ad the figures for atomic mass ad atomic umber. Use the otatio X A, where is the atomic umber, X is the elemet ame ad A is the atomic weight to the earest iteger. Accordig to the displacemet law of radioactivity, whe a atom udergoes alpha decay (i.e. helium uclei), the atomic weight decreases by 4 ad the atomic umber decreases by. Whe the atom udergoes beta decay (i.e. electros), the atomic weight does t chage ad the atomic umber icreases by. By applyig these rules ad usig a periodic table we get the followig series. 9U 35 α 9 Th 3 β 9 Pa 3 α 89 Ac 7 β 9Th 7 α 88 Ra 3 α 86 R 9 α 84 Po 5 α 8Pb β 83 Bi β 84 Po α 8 Pb 7 (stable isotope) 5. Carbo-4 datig. Carbo-4 is a radioactive isotope (half life of 573 years), created i the atmosphere by cosmic rays, that combies with oxyge to create carbo dioxide that is the absorbed by plats (ad evetually makes it ito plat eatig orgaisms). Eve though C 4 is always decayig, it is cotiually beig repleished i livig thigs so that the ratio of C 4 to ormal carbo (C ) is early costat. Oce the orgaism dies, the C 4 is o loger repleished ad simply decays away. This decay ca be used to determie how log ago the orgaism died. The dirt floor of the Shaidar Cave i the orther part of Iraq has bee examied. Below the layer of soil that cotaied arrowheads ad boe awls

3 was a layer of soil that yielded flit tools ad pieces of charcoal. Whe the charcoal was examied it was discovered that i kg of carbo, approximately 9.4x carbo-4 uclei decayed each secod. It is kow that i kg of carbo from livig material,.5x 4 disitegratios of carbo-4 occur each secod. Use this data to calculate whe people of the Stoe Age culture occupied the cave. Sice we are dealig with decays/sec, we ca use the expoetial formula for the decay rate (activity). Solvig for t, as i problem 3, ad pluggig i the umbers, we get 4 T/ R 573 years.5x decays / s t l l 898 years l R( l 9.4x decays / s 6. SMM, Chapter 3, problem 4. a (a) The radii of the various Bohr orbits are give by r where a. 59 m. Pluggig (hydroge) ad, & 3 we get r a.59 m r 4a.6 m r 3 9a.476 m (b) The electro s speed is obtaied by settig the Coulomb force equal to the cetripetal force (SMM eq. 3.6): v ke. m r 9 9 (8.988x Nm / C )(.6x C) v.9x 6 m/s.7c 3 9 (9.9x kg)(.59x m) v.9x 6 m/s.4c v 3 7.8x 5 m/s.c (c) The velocities are much smaller tha the speed of light. No relativistic correctio is ecessary. 7. SMM, Chapter 3, problem 5. (a) The eergy levels for a hydroge-like io whose charge umber is is give by E 3.6eV. For helium ( ), we plot a eergy level diagram similarly to SMM Figure 3.3, where I ve icluded the eergy level diagram of hydroge for compariso. e

4 (b) The ioizatio eergy is simply the eergy required to take the electro from the state to the state. For hydroge-like helium (He + ) this is just E 3.6eV ev. 8. SMM, Chapter 3, problem 6. We follow the same procedure as i problem 7 but for hydroge-like Lithium (Li + ) which has 3.

5 9. SMM, Chapter 3, problem 7. a Usig r (a) (Helium, He + ( ) a.65 m (b) 3 (Lithium, Li + ( ) a.76 m 3 (c) 4 (Beryllium, Be 3+ ( ) a.3 m 4 Due to the icreased positive charge i the ucleus, the first Bohr orbit gets smaller ad smaller.