An atom of tin has 50 protons, and (typically) 70, 68, or 66 neutrons. What are
ID: 1303568 • Letter: A
Question
An atom of tin has 50 protons, and (typically) 70, 68, or 66 neutrons. What are the symbols for the nuclei of these objects'? 14050Sn, 13650Sn 13250Sn 12070Sn, 11868Sn 11666Sn 12050Sn, 11850Sn 11650Sn 12030Sn, 11828Sn 11626Sn 5070Sn, 5068Sn, 5066Sn 7050Sn, 6850Sn, 6650Sn 2050Sn, 1850Sn, 1650Sn A nucleus of 4019K will sometimes capture the n=1 electron. What is the result of this capture? 4019K 4020Ca 4020Ca +0-1e 4018Ar 4118Ar 0-1e Carbon-14 is famously unstable, decaying spontaneously. What is wrong with this isotope? simply too big simply too small too many neutrons too many protons too many electrons too many unpaired nucleons What will carbon-14 do, to become more stable? capture a proton capture a neutron capture an electron capture a positron emit a proton emit a neutron emit an electron emit a positron it is already stable, so it won't do anything You break a neutron apart o a proton and an electron. What's the change in mass Energy during this reaction? {Did you need to add that Energy or was it emitted ?} mnc2 = 939.565[MeV]...mpc2 = 938.272[MeV]...mec2 = 0.511[MeV] capture an electron capture a positron emit a proton emit a neutron emit an electron emit a positron it is already stable, so it won't do anything You break a neutron apart o a proton and an electron. What's the change in mass Energy during this reaction? {Did you need to add that Energy, or was it emitted ?} mnc2 = 939.565[MeV]... mpc2 = 938.272[MeV]...meC2 = 0.511 [MeV] 1878.348[MeV]... added 1878.348{MeV]... released 1.804[MeV]. added 1.804[MeV] released 1.293(MeV]. added 1.293{MeV]. released 0.782[MeV] added 0.782[MeV] released What is the binding energy per nucleon for the nucleus 19779Au ? 197Au mass = 196.966 543 u ; 1H mass = 1.007 825 u ; 1n mass = 1.008 665 u ; ... use (1 u) c2 = 931.5 MeV (fig.29.4; is a graph of the negative PE... "how stable" each isobar is = how "deep"the PE gets at each A, across the valley of stability} 7.76 [MeV] 7.92 [MeV] 8.230 [MeV] 19.74 [MeV] 1559.9 (MeV] 1621.75 [MeV] A radioactive object is observed to have an activity of 1 000 decays/sec at noon, and six hours later it is observed to have an activity of 125 decays/sec. What is its half-life?1000/37000 = 0.0270 hour 1/3 hour 6/3 hour =1/2 hour 6/8 =3/4 hour 8/6 = 4/3 hour 6/3 =2 hour 6/2 =3 hour 6/0.875 = 6.86 hour 1/0.125 = 8hour 6/0.125 = 48 hour 6/8 = ¾ hour 8/6 = 4/3 hour 6/3 = 2 hour 6/2 = 3 hour 6/0.875 = 6.86 hour 1/0.125 = 8 hour 6/0.125 = 48 hour A throwing spear was found in Austria, and a 0.2 microgram sample (dry, .03mm x 03mm x .2mm) contained 2500 atoms of 14C. Knowing that 14C has a half-life of 5730 years, and presuming that such a sample would have originally contained 80,000 carbon-14 atoms (as growing trees did in the 1930's), how long ago did the spear-tree stop growing? 179 years 8270 years 28.700 years 183.000 years 367.000 years 104.000.000 years 4.29 times 109 years A pure sample of 226Ra contains 2.0 E14 atoms . If the half-life is 1600 [yr], what is the decay rate = "Activity" of this sample? {1 nano-Curie = 1 [nCi] = 37 [decays/s]} 2.34 nCi 74.2 nCi 86.6 nCi 107.0 nCi 154.5 nCi 101.6 1/4CI 118.7 1/4Ci Tritium (3H) has a half-life of 12.3 years and releases 0.018 6 MeV energy per decay. What is the rate at which energy is released for a 1.0-gram sample of tritium? (NA = 6.02 times 1023 mol-1, 1 year = 3.16 times 107 s, 1 MeV = 1.6 times 10-13 J, and mt = 3.016 05 u) 0.33 W 1.1 W 9.6 W 3.2 W Click Save and Submit to save and submit. Check Save All Answers to save all answers.Explanation / Answer
5) the reaction is
neutron ---> proton + electron
change in mass energy = mass energy of proton + mass energy of electron - mass energy of neutron
change in mass energy = 938.272 + 0.511 - 939.565
chnage in mass energy = -0.782 MeV
so change in mass energy = -0.782 MeV
here - sign indicates that enrgy is released so
the answer is 0.782 MeV , released
6) mass defect dm = 79 x mass of proton + 118 x mass of neutrons - mass of nucleus
dm = 79 x 1.007825 + 118 x 1.008665 - 196.966543
dm = 1.674 u
so binding energy = dm c2
binding energy = 1.674u x c2
binding energy = 1.674 x 931.5
binding enrgy = 1559.426 MeV
binding energy per nucleon = 1559.426 / 197
binding energy per nucleon = 7.915
so the binding energy per nucleon = 7.92 MeV
9) we know that
Activty A = lamda N
we know that lamda = 0.693 / t1/2
so lamda = 0.693 / 1600 x 365.25 x 24 x 60 x 60
lamda = 1.3724 x 10-11 s -1
given N = 2 x 1014
so A = 1.3724 x 10-11 x 2 x 1014
A = 2744.98 decay /s
given 1 nci = 37 decay /s
so A = 2744.98 / 37 nci
A = 74.188 nci
so activity is 74.2 nci