Calculating Radioactive Decay Honors Chemistry Dr. Yager.

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Presentation transcript:

Calculating Radioactive Decay Honors Chemistry Dr. Yager

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

M(t) = M o e -kt M(t) = M o e -kt Definition of half life (t 1/2 ): In one half life, half the mass has decayed, half is left. or at time t = t 1/2, M = M o /2 or at time t = t 1/2, M = M o /2 M(t) = M(t 1/2 ) = M o /2 = M o e -kt 1/2 M(t) = M(t 1/2 ) = M o /2 = M o e -kt 1/2 ½ = e -kt 1/2 ½ = e -kt 1/2 ln(½) = ln(e -kt 1/2 ) = -kt 1/2 ln(½) = ln(e -kt 1/2 ) = -kt 1/ = -kt 1/ = -kt 1/2 k = 0.693/t 1/2 k = 0.693/t 1/2 M(t) = M o e -kt = M o e -(0.693/t 1/2 )t M(t) = M o e -kt = M o e -(0.693/t 1/2 )t

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 = M o e -kt Use: M = M o e -kt so; k = (0.693/2.60 h) = h -1 M = 1.0 mg (e x 10.4 ) = mg

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) = days -1 k = (0.693/24.1 days) = days -1 M = 1.0 mg (e x48.2 ) = 0.25 mg M = 1.0 mg (e x48.2 ) = 0.25 mg

M(t) = M o e -kt M(t) = M o e -kt ln M = ln (M o e -kt ) ln M = ln (M o e -kt ) = ln M o + ln e -kt = ln M o + ln e -kt = ln M o - kt = ln M o - kt ( y = b + mx ) ( y = b + mx )

How many protons are in 1 mole of ? 1 mole of anything is equal to 6.02 x of those items. For example: 1 mole of atoms is 6.02x10 23 atoms 1 mole of cookies is 6.02 x cookies

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 - 6 protons + 8 neutrons = 14 nucleons; n → p + + e - The neutrons decay to protons; therefore, there are: (3.61x ) protons = 3.61 x protons (3.61x ) protons = 3.61 x protons half of the neutrons change to protons x new protons Total protons 6.02 x protons Total protons 6.02 x protons The new material formed is N-14. The new material formed is N-14.