Question

Calculate the activity due to 4C in 1 . 00 kg of
carbon found in a living organism . Express the activity in units
of Bq and Ci . It is known , that the half - life of 4C is 5730
years.

Answer #1

An archeological specimen containing 7.5 g of carbon has an
activity of 1.1 Bq. How old (in years) is the specimen? Carbon-14
has a half-life of 5730 years, and an activity of 0.23 Bq per gram
of carbon in a living organism.

In a living organism, a fixed fraction 1.30 x 10-12 of 12C is
the radioactive isotope 14C, which has a half life of 5730 y. What
is the activity of the 14C in .95g of 12C found in living tissue in
Bq?

44g of petrified wood was found in a petrified forest. A sample
showed a carbon-14 activity of 100 decays/minutes. How long has the
tree been dead (in years)? (The half-life of carbon-14 is 5730
years and freshly cut wood has a carbon-14 activity of 14.96
decays/min per gram.)

Carbon-14 is unstable and undergoes β- decay, with a
half-life of 5730 years.
(a) What does it decay into?
(b) How much energy is released in the reaction?
(c) If a 1 g sample of carbon contains 0.1% carbon-14, what is the
activity in units of curie (Ci), where 1 Ci
= 3.7 × 1010 decay/s?
(d) What is the activity after 20,000 years have passed?

An ancient club is found that contains 100 g of pure carbon and
has an activity of 6.5 decays per second. Determine its age
assuming that in living trees the ratio of (14C/12C) atoms is about
1.30×10^-12. Note that the half life of carbon-14 is 5700 years and
the Avogadro number is 6.02×10^23.
Ans in Years

An ancient club is found that contains 100 g of pure carbon and
has an activity of 6 decays per second. Determine its age assuming
that in living trees the ratio of 14C/12C
atoms is about 1.10 × 10-12. Note that
the half life of carbon-14 is 5700 years and the Avogadro number is
6.02 × 1023.

An ancient club is found that contains 180 g of pure carbon and
has an activity of 7.5 decays per second. Determine its age
assuming that in living trees the ratio of (14C12C)(14C12C) atoms
is about 1.30 × 10-12. Note that the
half life of carbon-14 is 5700 years and the Avogadro number is
6.02 × 1023.
____years

The cloth shroud from around a mummy is found to have a 14C
activity of 8.0 disintegrations per minute per gram of carbon as
compared with living organisms that undergo 16.3 disintegrations
per minute per gram of carbon. From the half-life for 14C decay,
5715 yr , calculate the age of the shroud. Express your answer
using two significant figures.

Part A
You are using a Geiger counter to measure the activity of a
radioactive substance over the course of several minutes. If the
reading of 400. counts has diminished to 100. counts after 78.4
minutes , what is the half-life of this substance?
Express your answer with the appropriate units.
Part B
An unknown radioactive substance has a half-life of 3.20 hours .
If 26.8 g of the substance is currently present, what mass
A0 was present 8.00 hours...

You are using a Geiger counter to measure the activity of a
radioactive substance over the course of several minutes. If the
reading of 400. counts has diminished to 100. counts after 54.3
minutes , what is the half-life of this substance?
Express your answer with the appropriate units.
t1/2 =
27.2 min
Correct
Part B
An unknown radioactive substance has a half-life of 3.20 hours .
If 20.6 g of the substance is currently present, what mass
A0 was...

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