Question

Radioactive​ uranium-235 has a​ half-life of about 700 million years. Suppose you find a rock and chemical analysis tells you that only one sixteenth
of the​ rock's original​ uranium-235 remains. How old is the​ rock?

A.
2.8 billion years old

B.
1.4 billion years old

C.
175 million years old

D.
2.1 billion years old

E.
3.5 billion years old

F.
350 million years old

Answer 1

Answer:

Option A.

Step-by-step explanation:

Half life of Uranium-235 has been given as 700 million years.

Since radioactive decay is an exponential phenomenon so the formula will be

$A_{t}=A_{0}e^{-kt}$

where $A_{t}$ = Amount of the radioactive element at the time 't'

$A_{0}$ = Initial amount

t = time for decay

k = decay constant

By this formula,

$\frac{A_{0}}{2}=A_{0}e^{-kt}$

$\frac{1}{2}=e^{-700\times 10^{6} k}$

By taking natural log on both the sides,

$ln(\frac{1}{2})=ln(e^{-700\times 10^{6}k } )$

$-ln2=-700\times 10^{6}\times k$

0.693147 = $700\times 10^{6}k$

k = $\frac{0.693147}{700\times 10^{6}}$

  = $\frac{0.693147}{7\times 10^{8}}$

  = $9.9\times 10^{-10}$

Now we have to find the age of the rock which is one sixteenth of the original rock.

By the formula again,

$A_{t}=A_{0}e^{-kt}$

$\frac{A_{0}}{16}=A_{0}e^{-kt}$

$\frac{1}{16}=e^{-9.9\times 10^{-10}t}$

Taking log on both the sides.

$ln\frac{1}{16}=ln(e^{-9.9\times 10^{-10}t})$

$2.772588=-9.9\times 10^{-10}\times t$

t = $\frac{2.772588}{9.9\times 10^{-10} }$

t = $0.28\times 10^{10}$

t = $2.8\times 10^{9}$

Therefore, the rock is 2.8 billion years old.

Option A. is the answer.

Answer 2

Answer:

  A.  2.8 billion years old

Step-by-step explanation:

1/16 = (1/2)^4, so 4 half-lives have elapsed.

  4 · 0.700 billion years = 2.8 billion years

The rock is 2.8 billion years old.