An atom of tin has 50 protons, and (typically) 70, 68, or 66 neutrons. What are
ID: 1303431 • 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 7050Sn,6850Sn,6650Sn, 2050Sn,1850Sn, 165DSn 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 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 0.782JMeV] ... added 0.782[MeV] ... released What is the binding energy per nucleon for the nucleus 19779Au 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]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
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.