Boron-12, an unstable nucleus, undergoes the following decay process. The half-l

Question

Boron-12, an unstable nucleus, undergoes the following decay process. The half-life for this beta decay is 20.2 milliseconds and the atomic mass of Boron-12 is 12.0143521 u (Anti-neutrino) electron 12 orray photon 12 12 a) What is the binding energy per nucleon for Boron-12? (5 pts) b) A 275 g sample of this isotope of Boron contains how many nuclei? (1 pt) c) If you started with a 275 g sample of Boron-12 at time 0, how many nuclei of this isotope would be left 0.500 seconds? (3 pts) d) What is the activity of this 275 g sample of Boron-12 in decays/second att 0.500 seconds? (I pt)

Explanation / Answer

a) number of protons - 5

Number of nuetrons - 7 (12-5)

Mass of 1 proton - 1.00728 amu

Mass of 1 nuetron- 1.00867

Combined mass is calculated as,

( 5* 1.00728)+ (7* 1.00867)=12.09709

Now we have to calculate mass defect

?m= 12.09709-12.0143521=.0827379 amu

Next we have to convert mass defect into energy

We know 1 amu= 1.6606* 10^-27 kg

So total mass defect in kg = .0827479*(1.6606*10^-27)

= 1.373945*10^-28 kg/nucleus

Convert this mass into energy by ?E=?mc^2

Where c= 2.9979*10^8 m/s

E=(1.373945*10^-28)*(2.9979*10^8)= 1.23482*10^-11 j/nucleus

To convert this into Me( mega electron volts) per nucleon,

1 Mev=1.602*10^-13 J

Then (1.23482*10^-11)/(1.602*10^-13 * 12)=6.4234 Mev/ nucleon

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