Question

question. Correct answers for each question will be made available once the maximum number of attempts have been made for each submission. Nuclear Physics - Binding Energy 1 Nuclear Physics - Binding Energy Fod work The above figure shows a nuclear reaction in which an alpha particle separates into individual nucleons. An a particle is a He nucleus (2 protons, 4 nucleons). The relevant rest masses are: He = 4.002602 u, neutron = 1.008665 u, proton = 1.007825 u. (a) What is the mass defect in atomic mass unit u? Keep 6 decimal places. Report the mass defect as positive. Enter a number u Submit (5 attempts remaining) (b) What is the energy (in MeV) corresponding to the mass defect? Keep 3 decimal places. (1uc? = 931.5 MeV) Enter a number MeV Submit (5 attempts remaining) Is energy released in this reaction or must energy be supplied for the reaction to occur? Select your answer from one of the following options. a. Energy is released, since the total rest energy before the decay is smaller than the total rest energy after the decay. b. Energy is supplied, since the total rest energy before the decay is smaller than the total rest energy after the decay. c. Energy is released, since the total rest energy before the decay is larger than the total rest energy after the decay. d. Energy is supplied, since the total rest energy before the decay is larger than the total rest energy after the decay. Submit (5 attempts remaining) Dividing the energy in (b) (which is the total energy involved in separating one nucleus) by the total number of nucleons, the result is called Binding Energy per Nucleon. (c) What is the value of Binding Energy per Nucleon for alpha particle? Keep 3 decimal places. Enter a number MeV pernucleon Submit (5 attempts remaining)

Answer 1

- Mreactant ma OL Given, he 9.00 2602 Alpha pare ficle Reparates in to nucleons .( X-te) a → 2 P + 20 mass defect, Am M product (2 mp +2mn)-ma =(2 x 1.0078250 + 2x 1.008665) mn=1.00 86654 4.002602 mp= 1.007825 e. 4.03298- 4.002602 0.0303784 O. OBOU Energy corresponds to mass defect 16 5 Am (b) Ame? 2 0.030378cech w duc = 931.5 Mev) 0303 78 x 931. 5 Mev 28.297107 Mev = 28,297 Mey * Ansb Energy is supplied since terkel rest energy energy aftere beforce the decay is smaller than total rest the decay © Binding energy pere nucleon, 28.297107 4 7. 0742 7675 A E 7.074 Mev perinucleon * la 0.693 tlun tur 2 Half life 18 given by, ti2 = 6408 days 102 (@) we know that , de cay constant, a= x = 0.010696 71 0.010 day converting into second, tlega 64.8 days = 64.6 X24*60x60 5698720 seconds 0693 5598720 seconda S 10237X10

case-lomass of initial readioachve sample = 6:lokg Decayed makrelal 0.440 kg. so material reem remaining 6:10 - 0.440 = 5,66kg Mass of readioactive 5.66 kg 18 sample that reemains we [N+= know that, -xt reemain! NA sempre at det No e uning 0.010t No = Inktal sample 5.66 6.10e 5.66 0.0lot 6:1 -0.0lot o.otot = 1.0777 0.9278 > e 0.9298 e > 0.010t= ln(1.0770) = 0·074829 > ta 0.074829 7,48 291.406 days 0:00 . The time is 7.48 days case-2 IN # / briven initial activity, Ao = 11263.82 decoup/s. Actvity, AndN at , AU NCE Nt relationship bet A and a A=XX N. No= number of nudei in initial sample 11263.82 decey's 9105.75 x 10 1.237x10-7 g 9.10x1olo Ido Activity in decayls after a days Activity at time t, Alt = Age a -xt - A 7 1.237X 10 67x6048oo = 11263.82x e 10u5l. 882 de causs - Az