ch43 kho tài liệu bách khoa
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ch43 kho tài liệu bách khoa
1If/? is the fission rate, then the power output is /’ RQ. where Q is the energy released in each fission event. Hence,R = P!Q = (1.0 W).'(200 X 1 o6 ch43 kho tài liệu bách khoa cV)( 1.60 X 10 19 J/cV) = 3.1 X 1010 fissions/s.2We note that the sum of superscripts (mass numbers A) must balance, as well as the sum of z values (where reference to Appendix F or G is helpful). A neutron has z = 0 and .4=1. Uranium has z= 92.(a)Since xenon has z = 54. then “Y” must have z - 92 - ch43 kho tài liệu bách khoa 54 - 38. which indicates the clement Strontium. The mass number of “Y” is 235 + 1 - 140 - 1 = 95, so “Y” is *>?Sr.(b)Iodine has z - 53, so “Y“ has zch43 kho tài liệu bách khoa
- 92 - 53 - 39. corresponding to the clement Yttrium (the symbol for which, coincidentally, is Y). Since 235 + I - 139 - 2 = 95. then the unknown isot1If/? is the fission rate, then the power output is /’ RQ. where Q is the energy released in each fission event. Hence,R = P!Q = (1.0 W).'(200 X 1 o6 ch43 kho tài liệu bách khoais 235 + I - 100 - 2 - 134. so we obtain |UTc.(d)Examining the mass numbers, we find /> 235 + 1 - 141 - 92 - 3.3(a) The mass of a single atom of * ’5U is (235 u)( 1.661 X 10 2 kg/u) = 3.90 X 10" 25 kg, so the number of atoms in 1.0 kg is(1.0 kg)/(3.90 X 10 2 ■ kg) 2.56 X Io24 2.6 X I o24.An alternat ch43 kho tài liệu bách khoae approach (but essentially the same once the connection between the *’u” unit and Nạ is made) would be to adapt F.q. 42-21.(b)The energy released bych43 kho tài liệu bách khoa
A' fission events is given by E = NO. where Ọ is the energy released in each event. For 1.0 kg of 235u,£=(2.56x 1024)(200x 10* cV)(1.60 X 10 19 J/eV)=1If/? is the fission rate, then the power output is /’ RQ. where Q is the energy released in each fission event. Hence,R = P!Q = (1.0 W).'(200 X 1 o6 ch43 kho tài liệu bách khoa factor 3.156 X 107 s/y is used to obtain the last result.4Adapting Eq. 42-21. there areI J?0?8 .i(6-02XIO23/mol) = 2JxIO*4 ' 239 g/tnol /AL = NA =* -v,-uplutonium nuclei in the sample. If they all fission (each releasing 180 MeV), then the total energy release is 4.54 X 102<’ MeV.5if Afcr is the ma ch43 kho tài liệu bách khoass of a 52Cr nucleus and A/mb is the mass of a 26Mg nucleus, then the disintegration energy isQ - (MCr - 2A/Mg)c?2 - [51.94051 u - 2(25.98259 u)](931.ch43 kho tài liệu bách khoa
5 MeV/u) - - 23.0 MeV.6(a) We consider the process ”.Mo —> x 'Sc+ *Sc. rhe disintegration energy isQ = (mM0 - Zmsck2 = [97.90541 u - 2(48.95002 u)](931If/? is the fission rate, then the power output is /’ RQ. where Q is the energy released in each fission event. Hence,R = P!Q = (1.0 W).'(200 X 1 o6 ch43 kho tài liệu bách khoaously fissioning; the energy barrier (see Fig. 43-3) is presumably higher and/or broader for Molybdenum than for Uranium.1If/? is the fission rate, then the power output is /’ RQ. where Q is the energy released in each fission event. Hence,R = P!Q = (1.0 W).'(200 X 1 o6Gọi ngay
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