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Calculating Radioactive Decay

Calculating Radioactive Decay. Honors Chemistry Dr. Yager. The basic concept: At any given time t, the rate of change of mass of radioisotope is proportional to the mass of the radioisotope. Let M(t) = radioisotope mass at time t M o = M(0) = initial mass at t = 0

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Calculating Radioactive Decay

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  1. Calculating Radioactive Decay Honors Chemistry Dr. Yager

  2. The basic concept: At any given time t, the rate of change of mass of radioisotope is proportional to the mass of the radioisotope. Let M(t) = radioisotope mass at time t Mo = M(0) = initial mass at t = 0 k = constant From calculus, we can write dM(t)/dt = k M(t) (rate of change of mass is proportional to mass) which has the solution: M(t) = Mo e-kt

  3. M(t) = Mo e-kt Definition of half life (t1/2): In one half life, half the mass has decayed, half is left. or at time t = t1/2, M = Mo/2 M(t) = M(t1/2) = Mo/2 = Mo e-kt1/2 ½ = e-kt1/2 ln(½) = ln(e-kt1/2 ) = -kt1/2 -0.693 = -kt1/2 k = 0.693/t1/2 M(t) = Mo e-kt = Mo e-(0.693/t1/2)t

  4. Manganese-56 is a beta emitter with a half-life of 2.60 h. What is the mass of a 1.0 mg sample of Mn-56 after 10.4 h? Use: M = Moe-kt so; k = (0.693/2.60 h)= 0.266 h-1 M = 1.0 mg (e-0.266 x 10.4) = 0.063 mg

  5. Thorium-234 has a half-life of 24.1 days. How much of a 1.0 mg sample will be left after 48.2 days? k = (0.693/24.1 days) = 0.0290 days-1 M = 1.0 mg (e-0.0290x48.2) = 0.25 mg

  6. M(t) = Mo e-kt ln M = ln (Mo e-kt) = ln Mo + ln e-kt = ln Mo - kt ( y = b + mx )

  7. How many protons are in 1 mole of ? 1 mole of anything is equal to 6.02 x 1023 of those items. For example: 1 mole of atoms is 6.02x1023 atoms 1 mole of cookies is 6.02 x 1023 cookies

  8. How many protons are left after one half-life of β decay of one mole of C-14 ? 6 protons+ 8 neutrons= 14 nucleons; n → p+ + e- The neutrons decay to protons; therefore, there are: (3.61x 1024) protons = 3.61 x 1024 protons half of the neutrons change to protons+ 2.41 x 1024 new protons Total protons 6.02 x 1024 protons The new material formed is N-14.

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