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November 23, 2014

November 23, 2014

Posted by **bryce tanaka** on Friday, October 7, 2011 at 10:57am.

Because the calculations involve the small difference of (comparatively) large numbers, you need to keep 7 significant figures in your calculations, and you need to use the more accurate value for the speed of light, 2.99792e8 m/s.

Choose all particles as the system.

Initial state: Original nucleus, at rest.

Final state: Alpha particle + new nucleus, far from each other.

What is the rest energy of the original nucleus? Give 7 significant figures.

What is the sum of the rest energies of the alpha particle and the new nucleus? Give 7 significant figures.

- Physics -
**drwls**, Friday, October 7, 2011 at 11:20amCalculate the loss of rest mass energy. That will equal the sum of the kinetic energies of the two fragments.

(delta m)*c^2 = 9.322*10^-30 kg*c^2

= 8.379*10^-13 J

Since the original particle was stationary, the two fragments have equal and opposite momenta.

You can use that fact and the total final kinetic energy to get the velocites of each of the two fragments. Hoever, all thay ask you to do is compute m c^2 for each partricle and compare initial and final values. That has already been done above.

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