Question

If the average energy released in a fission event is 208 MeV, find the total number of fission events required to operate a 120-W lightbulb for 1.5 h.

Answer 1

The number of fission are equal to 1.947 x 10¹⁶

Step-by-step explanation:

Energy of one fission=208*1.6*10⁻¹³ Joule.

Power of the bulb=120 Watts.

Time t in seconds = 1.5h=5400 seconds

Total energy E=Power* time=120*5400=64.8 x 10⁴ Joule.

Number of fissions N= total energy / energy for one fission

=(64.8 x 10⁴)/(208x1.6x10⁻¹³)

=1.947 x 10¹⁶