Question

Questions 1-4

1.Calculate the mass defect and the binding energy/nucleon of the nuclide Be, which has a mass of 9.012 182 24 amu. The mass of a proton is 1.007 276 47 amu and the mass of a neutron is 1.008 664 90. One amu - 1.6605 x 10 kg and the speed of light is 3.00 x 10 m/s. 2. Write the nuclear equation for each of the following reactions. Refer to a periodic table. a. the alpha decay of Ra b. the beta decay of ci c. the positron emission of sp 3 Phosphorus-32 has a half-life of 14.3 days. How many milligrams of phosphorus-32 remain after 71 days if you start with 4.00 mg of the isotope? 4. What is the half-life of an isotope if after 2.00 weeks you have 31.25 g remaining from a 250.0 g starting sample size?

Answer 1

Mass num ber Be Atomic number num ber 4 Atomic Num bey of Probons Atomic numser Mass num ber- Num bey of neuhons contains 5 neutrons Be 4 pro lonA mass pone proton= .o0 7 276 47 4m The one neutbon 100 G6y 90 amy mass of The mas op 4 pro fons 4xoo7 27647amy Now L4.02 910588 a.m.4. s neutbons 5xoo 9 Gey 90 am-y 5.0433 245 am.y Total mas amy (4.029105 88+ 5.a4 332 9.07-24 30 38 a.m.y 9.012 1P2 2yamy Actuel moss ep F wed wad macr CAm) Actr Catrdstelmag Mass dep e c Caleulated Actual mass Am 9.012182 24)amy Ca. 24303 O.0 6024 8 14 am.4 Den 27 27 l0004 20 365x1o

Binding energy anucleus C be caleulated EinStein' equation E eplere +ese Arn lightvelouty 3 x10mls einding ene3y Reduced mao lass deprct 7 .100o 420 36sx(D C31OMIS E D 1o0o420 3 65 x10 ka 27 o.000420 365 xo x 3 IG O.l000 420 365 xIo 7+16 o.900378 32-82 > lo O.900 378 3282 xIO o.4003783282 x jo Tulea Ik3=1oooTulex Ktlo Touler I Toule kT -3 KT o.4a0 318 32. 82 X10 xio K 7 E KJ O.900 378 32 2x1o ele cheon Valta cev) 21 ev 6.242 x lo IkT 21 E-0.4o0 3783282 xto x G.242x10 ev to 5.6201615248X10 eV -3 to 5.6201G15 24&x ix lo kev

E 5.62016152y8X(0 kev equationstox Nude Pellocwing reachon 22 The decay op Ra 88 He x putic le the a-paur hicle emits Pro S0 when an nucleuS atomic num er and ma5 number op decrease s by 24 units especivey etement an a22 Ra He Sc 2 Element orth atomic numsor 86 iA Radon 226 Equation Fer decoy op Ra et desntegaation22 22-6 Ra -tHe 2 pasn bi cle 39 ihe F-decay 17 Ppar hcle e p-panticle emits prom the when Co nucleus mass num ber element an does not change but atomic nu ber is t0re aned one uni t 29+0 39 e cl 17+1 tomic number 18 i Avgon o , th e lemant An CA Equation 1 39 pdisinttgraion Ar + ppanticle 17

posi tren emmi scron p C The Pastaon +ve ly changed p panticle So, the emmission pooton aesulta 1n the atomic num bes by decoea se op one unit change inthe mass 30 +0 P + e +1 positron 1S 14 e lement with atomic nunber 4 iA S An Equation op positaon emmi ssion 30 30 p + e 14 poohon 3 Initial amount 14.3 day s a 7 days Totl Hme op halp lives hale lire 7 days buprhi o.489510489 the amaud op Subtve tptrle halt livex ()-

o.4895104895 4.20mg o.48 95104 895 2 x4-00m9 140 396 8 y246 9-8490 66 923 67my Amont wemain ingaf ter 7days 2.8490 66 12167m If you hve ese ven total days ct other than dyyou Can t 7 by -the given num ber. 950. 09r 8a Inial mount 4 Amount emainin 31 25 S ee ka 14 days Trme kno the tale Initiel amotat Amant yernai niny an beth side Tang logaat thn

n lagio (0.5) Total time hese CANoopbale liver 1ang tys L03 0.5 Ry be'subatituting the value t, 2 Ga ve can get t2 a Pollou H days l0g (0-5) fo /31-257 14 days x l0310 C0.125) o.30/02.999566 14 dayx .90 30 8998679 14 days X o.333 33 4.666 dayA Ie4664 Jays