Question

Estimate how much fission must occur each second to supply a day's energy to a private household in a temperate climate (about 55 kilowatt-hrs). Assume each fission releases 200 MeV (million electron volts). Use ® 1 eV = 1.6 x 10-19J and/or 1 watt = 1 J/s

Answer 1

To supply a day's energy to a household (55 kWh), approximately 7.16 x 10^13 fissions must occur each second, assuming each fission releases 200 MeV.

Step-by-step explanation:

To estimate the amount of fission required per second to supply a day's energy to a private household in a temperate climate (about 55 kilowatt-hours), we will first convert this energy requirement into joules using the conversion 1 watt-hour = 3600 joules:

55 kWh * 3600 J/Wh = 198,000,000 J (198 MJ)

Next, we need to convert the energy released per fission event into joules:

200 MeV/fission * 1.6 x 10-19 J/eV = 3.2 x 10-11 J/fission

Now, we can calculate the total number of fissions needed to produce 198 MJ:

198 MJ / (3.2 x 10-11 J/fission) = 6.1875 x 1018 fissions

Since this is the total number of fissions needed for one day, to find the number of fissions per second, we divide by the number of seconds in a day (86,400 seconds):

6.1875 x 1018 fissions / 86,400 s = 7.16 x 1013 fissions per second

Therefore, to supply a day's energy to a household, approximately 7.16 x 1013 fissions must occur each second.