Radioactive uranium‐238 decays sequentially into thirteen other lighter elements until it stabilizes at lead‐206. The half‐lives of the fifteen different elements in this decay chain vary from 0.000 164 seconds (from polonium‐214 to lead‐210) all the way up to 4.47 billion years (from uranium‐238 to thorium‐234).

Complete question is;

Lead- 206 is not radioactive, so it does not spontaneously decay into lighter elements. Radioactive elements heavier than lead undergo a series of decays, each time changing from a heavier element into a lighter or more stable one. Eventually, the element decays into lead-206 and the process stops. So, over billions of years, the amount of lead in the universe has increased because of the decay of numerous radioactive elements produced by supernova explosions. Radioactive uranium- 238 decays sequentially into thirteen other lighter elements until it stabilizes at lead- 206. The half-lives of the fifteen different elements in this decay chain vary from 0.000 164 seconds (from polonium- 214 to lead- 210 ) a|| the way up to 4.47 billion years (from uranium- 238 to thorium- 234 ). Find the decay rate per billion years for uranium- 238 to decay into thorium- 234 . Round your answer to one decimal place.

0.14363

Step-by-step explanation:

We are told that time could be all the way up to 4.47 billion years.

Now, formula for half life exponential equation is;

N = N_o•e^(λt)

In this case, N = ½N_o

Thus;

½N_o = N_o•e^(-λt)

N_o will cancel out to give;

½ = e^(-λt)

In ½ = -λt

-0.6931 = -λt

Putting 4.47 for t, we have;

-0.6931 = -4.47λ

λ = 0.6931/4.47

λ = 0.155056

Now, formula for decay rate constant is;

N_t = N_o(e^(-λt))

Since we want to find the decay rate per billion years, then t = 1.

Thus;

Putting 0.155056 for λ and 1 for t, we have;

N_t = N_o(e^(-0.155056 × 1))

N_t = N_o(0.85637)

N_t/N_o = 0.85637

Thus decay rate per 1 billion years is;

1 - 0.85637 = 0.14363

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step-by-step explanation:

i think it will be zero as slope has a formula of difference between initial and final distance divided by itz time interval