then what is the half-life for this radioactive element (i.e., how long will it take for 10 grams to reduce to 5 grams)?

Question

If 21 grams of radioactive waste reduces to 8.75 grams of radioactive waste after 1400 years, then what is the half-life for this radioactive element (i.e., how long will it take for 10 grams to reduce to 5 grams)? An exponential model is 𝑦 = 𝑦𝑜𝑒^ 𝑘𝑡 .

Answer 1

1400 * ln(2)/ln(21/8.75) = 1400 * ln(2)/ln(84/31) = 973.492418640262

Note that 2^-(1400/973.492418640262) = 31/84 = 8.75/21

Answer 2

y = y₀e^kt

8.75 = 21e^1400k

8.75/21 = e^1400k

ln(8.75/21) = ln(e^1400k)

ln(8.75/21) = 1400k•ln(e)

ln(8.75/21) = 1400k

k = ln(8.75/21)/1400 ≈ −0.000625335

So y = y₀e^{−0.000625335t}

To find the half life, Let y = 1/2y₀ then

1/2y₀ = y₀e^{−0.000625335t}

1/2y₀/y₀ = e^{−0.000625335t}

1/2 = e^{−0.000625335t}

ln(1/2) = ln(e^{−0.000625335t})

ln(1/2) = (−0.000625335t)ln(e)

ln(1/2) = −0.000625335t

t = ln(1/2)/−0.000625335 = 1108.4 years