Biology, asked by quadraj967, 11 months ago

The half-life for the radioactive decay of C-14 is 5730 years.How long will it take for 30% of the C-14 atoms in a sample of C-14 to decay?

Answers

Answered by Jyotimodi
0

Answer:

The time is

=

2948.5

y

Explanation:

The half life of

C

14

is

t

1

2

=

5730

y

The radioactive constant is

λ

=

ln

2

t

1

2

=

ln

2

5730

=

1.21

10

4

y

1

Apply the equation

m

(

t

)

=

m

0

e

λ

t

m

(

t

)

m

0

=

e

λ

t

λ

t

=

ln

(

m

(

t

)

m

0

)

t

=

1

λ

ln

(

m

(

t

)

m

0

)

The time is

t

=

1

1.21

10

4

ln

(

0.7

)

=

2948.5

y

Answered by Riya1045
0

Explanation:

It is given:

Charcoal has the initial activity, which is denoted as A_{0}=15.3A

0

=15.3 disintegrations per minute per gram.

Charcoal has the half-life, T 12=5730T12=5730 years

After a few years, the charcoal’s final activity, A = 12.3 disintegrations per minute per gram

Constant of disintegration,

\lambda=0.693 \mathrm{T} 12=0.6935370 \mathrm{y}-1λ=0.693T12=0.6935370y−1

For the action to attain 12.3 disintegrations per minute per gram, let the time taken by the sample at a time of t year.

Sample’s activity,

A=A O e-\lambda tA=AOe−λt

A=A O e-0.6935730 \times tA=AOe−0.6935730×t

\Rightarrow \mathrm{t}=1804.3 \text { years }⇒t=1804.3 years

Similar questions