An atom of tin has 50 protons, and (typically) 70, 68, or 66 neutrons. What are

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

Navigate

• Browse (All)
• Browse A
• Subjects

---