Question

The half-life of carbon-14 is about 5700 years. you have found a fossil that you believe to be about 34,200 years old because it has _____ the normal (modern) expected ratio of carbon-14c

Answer 1

Half-life of carbon-14 = 5700 years (given)

Half life is defined as the time needed for a amount to reduce to its half of its primary value. Here, carbon-14 is an radioactive isotope having 8 neutrons and 6 protons.

Assume, the value of carbon-14 is $x$.

After 5700 years, it will reduce to half and becomes= $\frac{x}{2}$.

Now, n (number of half-lives) is given by the ratio of elapsed time to the half-life.

$n =\frac{elapsed time}{half-life}$

Put the given values, we get:

$n =\frac{34200 years}{5700 years}$

= $6$

Now, the denominator comes out to be $2^{6}$ i.e. $\frac{1}{64}$ times the normal expected ratio of carbon-14.

Hence, the half-life of carbon-14 is about 5700 years. you have found a fossil that you believe to be about 34,200 years old because it has $\frac{1}{64}$ the normal (modern) expected ratio of carbon-14.

Answer 2

The 34200-years old fossil fuel has $\boxed{\frac{1}{{64}}}$ the normal (modern) expected ratio of carbon-14 C.

Further Explanation:

Radioactivity is the phenomenon due to which energy is released by an unstable atomic nucleus. This energy is in the form of various particles such as alpha particles, beta particles, and gamma particles.

Half-life

The time after which radioactive substance remains half of its original value is called half-life. It is denoted by ${{\text{t}}_{{\text{1/2}}}}$.

The exponential expression for half-life is as follows:

${{\text{A}}_{\text{t}}} = {{\text{A}}_{\text{0}}}{\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right)^{\left( {\dfrac{{\text{t}}}{{{{\text{t}}_{{\text{1/2}}}}}}} \right)}}$        …… (1)                                                          

Here,

${{\text{A}}_{\text{t}}}$ is the final amount of fuel.

${{\text{A}}_{\text{0}}}$ is the initial amount of fuel.

t is the time or age of fuel.

${{\text{t}}_{{\text{1/2}}}}$ is the half-life period.

Rearrange equation (1) as follows:

$\dfrac{{{{\text{A}}_{\text{t}}}}}{{{{\text{A}}_{\text{0}}}}}= {\left( {\dfrac{{\text{1}}}{{\text{2}}}}\right)^{\left( {\dfrac{{\text{t}}}{{{{\text{t}}_{{\text{1/2}}}}}}} \right)}}$                                                                    …… (2)

The value of t is 34200 years.

The value of ${{\text{t}}_{{\text{1/2}}}}$ is 5700 years.

Substitute 34200 years for t and 5700 years for ${{\text{t}}_{{\text{1/2}}}}$ in equation (2).

 $\begin{aligned}\frac{{{{\text{A}}_{\text{t}}}}}{{{{\text{A}}_{\text{0}}}}}&= {\left( {\frac{{\text{1}}}{{\text{2}}}}\right)^{\left( {\frac{{{\text{34200 years}}}}{{{\text{5700 years}}}}}\right)}}\\&= {\left({\frac{{\text{1}}}{{\text{2}}}} \right)^{\left( {\text{6}} \right)}} \\&=\frac{1}{{64}}\\\end{aligned}$

Therefore the required ratio of fossil fuel is 1:64.

Learn more:

1. What nuclide will be produced in the given reaction? https://brainly.com/question/3433940
2. Calculate the nuclear binding energy: https://brainly.com/question/5822604

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Nuclear chemistry

Keywords: half-life, alpha particles, beta particles, gamma particles, t1/2, half, original, radioactivity, energy, unstable atomic nucleus, time, 34200 years, 5700 years, 1:64, ratio, phenomenon, ratio, fossil fuel.