Calculating Energy Released from Uranium Fission

[estiy]

Homework Statement

a mass of 3.33x10^-28kg of uranium is converted into energy during nuclear fission (exploding) according to einstien's formula

E=mc^2
where energy is measured in joules (J), m is measured in kg and c = 3000000000m/s is the speed of light

Homework Equations

show that the amount of energy released during fission is 2.997x10^-1 J

The Attempt at a Solution

E = mc^2
= 28kg x (3x10^8ms^-1)^2
= 28kg x (3x10^8ms^-1)x(3x10^8ms^-1)
= 28kg (9x10^16m^2s^-2)
= 28 x (9x10^16)kg m^2s^-2
= 252 x 10^16 J
= 2,520,000,000,000,000

thats my attempt.. i know nothing about this stuff and i was trying to do this for a friend.. I am curious now what others come up with

 

[Gregg]

[itex] m = 3.33\times 10^{-28} [/itex]

[itex] c = 3\times 10^8 [/itex]

[itex] E = mc^2[/itex]

[itex] E = (3.33\times 10^{-28})(3\times 10^8)^2 = 2.997\times 10^{-11} [/itex]

 

[thomate1]

I would like to point out that your calculation is incorrect. The mass given in the problem is 3.33x10^-28kg, not 28kg. Also, the speed of light is 3x10^8m/s, not 3000000000m/s. So the correct calculation would be:

E = (3.33x10^-28kg) x (3x10^8m/s)^2
= 3.33x10^-28kg x (9x10^16m^2/s^2)
= 3.33 x (9x10^-12)kg m^2/s^2
= 2.997x10^-3 J

This is the amount of energy released during fission, which is 2.997x10^-3 J, not 2.997x10^-1 J as stated in the problem. It is important to carefully read and understand the given information in a problem before attempting to solve it.

Additionally, as a scientist, I would like to point out that the equation E=mc^2 is only applicable for mass-energy equivalence in special relativity, and may not accurately represent the energy released during nuclear fission. There are other factors and equations involved in calculating the energy released from nuclear fission, such as the binding energy per nucleon and the mass defect. It is important to have a thorough understanding of the physics and principles behind a problem before attempting to solve it.