Thorium-232 decays by a 10-step series of nuclear reactions, ultimately yielding

Question

Thorium-232 decays by a 10-step series of nuclear reactions, ultimately yielding lead-208, along with 6 ? particles and 4 ? particles.

I really need an in-depth explanation... Thank-you so very much!!!

Thorium-232 decays by a 10-step series of nuclear reactions, ultimately yielding lead-208, along with 6 ? particles and 4 ? particles. How much energy ( in kJ/mol ) is released during the overall process? The relevant masses are: 232Th Mu. He = 4.002603 u u, electron = 0.0005486 u Pb = 207.976627 u Th = 232.038054 u ; 208 Bb

Explanation / Answer

Since your question does't clarify about number of alpha and beta particle so only give guidline and you may do calculations your self

step1

determine mass fefect=(sum of masses of all product particles)-mass of thorium

step2

use E=mc2 where m is mass defect and c is speed of light c=3.0E8m/s

step3

the above may give energy in Jouls and amu units it may be converted to mole units

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