# Homework Help: Nuclear Energy

1. May 6, 2006

### proudtobeavol

I'm seriously lost!

Here's the problem:
It has been estimated that the Earth contains 1.0 x10^9 tons of natural uranium that can be mined economically. If all the world's energy needs (7.0 x10^12 J/s) were supplied by 235(U) fission, how long would this supply last? Assume that the average energy released in a fission event is 208 MeV. (Hint: See Appendix B for the percent abundance of _92^235(U).)

My (pathetic attempt at a) solution

OK, I've converted to 343000 tons of usable 235U and that this is equal to 3.11x10^12g which, using avogadro's number equals 7.97x10^33 nuclei available for a reaction. this converts to 1.65x10^36MeV which converts to 5.33x10^32 ev/kg

Now I need to get from here to how many years this would last. I tried multiplying the number of kg on hand (3111643658) by 5.33x10^32 ev/kg which gives me 1.66x10^42ev and since there are 6.24x10^18 J in an ev this gives me 2.66x10^23J

BUT I tried to go from there to 1s/7x10^12J=3.80x10^10s which converted to 1204 years... but it's wrong.

Does anyone see where I went wrong?

Much appreciation!
Shanna
(Who is hugely grateful that this is the LAST physics problem I will ever have to work!)

2. May 7, 2006

### jbusc

You intersperse grams with kilograms...plus you may have transposed some digits...so you're off by factors of 10.

Consider, you get 5.33x10^32 ev/kg ... but it is really 5.33x10^23 ev/g, which is 5.33x10^26 ev/kg.