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Thorium isotope (Th^232) has a half-life of 1.4*10¹⁰ years.

a)write the radioactive decay equation.
b) find the radioactive radiation activity of 10 g of thorium
c)and the activity after 10 years

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User Kerel
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a) The radioactive decay equation for Thorium-232 is:

Th-232 --> He-4 + Ra-228

b) The radioactive decay constant (λ) for Thorium-232 can be calculated using the half-life formula:

t1/2 = (ln 2) / λ

Rearranging the formula, we get:

λ = (ln 2) / t1/2

λ = (ln 2) / 1.4*10^10 years = 4.95*10^-11 per year

The radioactive radiation activity (A) of 10 g of Thorium-232 can be calculated using the following formula:

A = λ * N

Where N is the number of atoms of Thorium-232 in 10 g of Thorium. The number of atoms can be calculated using Avogadro's number (6.02*10^23 atoms per mole) and the molar mass of Thorium-232 (232 g per mole):

N = (10 g / 232 g per mole) * (6.02*10^23 atoms per mole) = 2.60*10^22 atoms

Therefore, the radioactive radiation activity is:

A = 4.95*10^-11 per year * 2.60*10^22 atoms = 1.29*10^12 decays per second

c) After 10 years, the activity of Thorium-232 will decrease due to radioactive decay. The remaining activity can be calculated using the following formula:

A = A0 * e^(-λt)

Where A0 is the initial activity, λ is the decay constant, t is the time elapsed, and e is the base of the natural logarithm (2.718).

The initial activity (A0) is the activity calculated in part b:

A0 = 1.29*10^12 decays per second

The time elapsed (t) is 10 years.

Therefore, the remaining activity is:

A = A0 * e^(-λt) = 1.29*10^12 decays per second * e^(-4.95*10^-11 per year * 10 years) = 1.09*10^12 decays per second

Therefore, the activity after 10 years is 1.09*10^12 decays per second.
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User Pbanka
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