al phy nuclear fission

发布网友发布时间：2023-09-17 02:10

共1个回答

热心网友时间：2023-09-22 22:00

更新1:

sorry
I made a typing mistake. Your first correction is correct but the "mass trfered into energy" should be calculated.

It seems that there are o typos in your question. The first one is that the power proced by the nuclear power station should be 900 MW(e) (900 MeV is a very very *** all power
equivalent to 10^-10 J. Thus 900 MeV is definitely a wrong figure. 900 MW(e) is similar to the power generated by a reactor at the Daya Bay Nuclear Power Station). The second typo is to calculate the "mass of U-235 atoms" under fission in one hour. Judging on the order of magnitude in the given options (which are quite large masses)
it would not be the "mass that trferred into energy". With the above corrections
the calculation is as follows: Thermal power proced by the power station = 900/0.4 MW = 2250 MW Thermal energy proced in one hour = 2250 x 3600 MJ = 8.1 x 10^6 MJ = 8.1x10^6/1.6x10^-19 MeV = 5.0625x10^25 MeV No. of U-235 atoms that undergo fission = 5.0625x10^25/200 = 2.531x10^23 Since 235 g ( 1 mole) of U-235 contain 6x10^23 atoms (Avogadro's number) hence
mass of U-235 that undergo fission = (2.531x10^23/6x10^23) x 235 g = 99 g = 0.099 kg Option C is the wer. 2012-04-05 20:16:35 补充： The mass trferred into energy = 8.1x10^12/(3x10^8)^2 kg = 9x10^-5 kg (using E = mc^2) The wer is option A. Sorry that I have mis-read the question.