Question 1

If we start with 1.000 g of strontium-90,0.908 g will remain after 4.00 yr.This means that the half-life of strontium-90 is ________ yr.

Accepted Answer

The half-life of a radioactive substance is the amount of time it takes for half of the original sample to decay. In this case, we can use the formula:

ln(Nt/N0) = -kt

Where Nt is the amount remaining after time t, N0 is the initial amount, k is the decay constant, and t is time. Solving for half-life (t1/2), we get:

t1/2 = ln(2)/k

We are given that Nt/N0 = 0.908/1 = 0.5 (since half of the original sample remains), and t = 4 years. Substituting these values into the formula above and solving for k, we get:

k = ln(2)/(4 t1/2) = ln(2)/(4 t1/2)

Solving for t1/2, we get:

t1/2 = ln(2)/(k*4) = ln(2)/(0.0883*4) = 28.8 years

Therefore, the half-life of strontium-90 is 28.8 years, and the best choice is C.

Question 2

The following reaction represents what nuclear process?  H +  H →  He +  n

Accepted Answer

This reaction represents nuclear fusion, where two hydrogen nuclei (protons) combine to form a helium nucleus and release a neutron.

Question 3

What percentage of a radioactive substance remains after 6.00 half-lives have elapsed?

Accepted Answer

After each half-life, the amount of a radioactive substance is reduced by half. Therefore, after 6 half-lives, the remaining amount is (1/2)^6 = 1/64 = 0.015625, which is 1.56%.

Question 4

Potassium-40 decays to argon-40 with a half-life of 1.27 × 10⁹ yr.The age of a mineral sample that has a mass ratio of ⁴⁰Ar to ⁴⁰K of 0.812 is ________ yr.

Accepted Answer

The answer of Potassium-40 decays to argon-40 with a half-life...

Question 5

Atoms with Z > ________ are radioactive and decay in one or more steps involving mostly alpha and beta decay.

Accepted Answer

Atoms with Z greater than 83 are unstable and radioactive. They undergo alpha and beta decay to achieve stability. Choice A (Z = 72) has stable isotopes, and choice B (Z = 130) has only one stable isotope. Choice D (Z = 150) has no stable isotopes. Choice E (Z = 30) has multiple stable isotopes.

Question 6

Strontium-90 is a byproduct in nuclear reactors fueled by the radioisotope uranium-235.The half-life of strontium-90 is 28.8 yr.What percentage of a strontium-90 sample remains after 75.0 yr?

Accepted Answer

The answer of Strontium-90 is a byproduct in nuclear reactors...

Question 7

Give the maximum age that can be estimated from radiocarbon dating.

Accepted Answer

Radiocarbon dating can only be used to estimate the ages of organic materials up to about 50,000 years old. This is because the half-life of the carbon-14 isotope used in radiocarbon dating is about 5,700 years, meaning that after this time, only half of the carbon-14 atoms in a sample would remain. Beyond about 50,000 years, the amount of carbon-14 left in a sample would be too small to accurately measure, making radiocarbon dating ineffective for estimating older ages.

Question 8

The following reaction represents what nuclear process?  U +  n →  Ba +  Kr + 3  n

Accepted Answer

This reaction represents nuclear fission, where the uranium-235 nucleus absorbs a neutron and splits into two smaller nuclei of barium-144 and krypton-89, with the release of three neutrons.

Question 9

Carbon-11 is used in medical imaging.The half-life of this radioisotope is 20.4 min.What percentage of a sample remains after 60.0 min?

Accepted Answer

After 60.0 minutes, which is approximately 3 half-lives of Carbon-11 (20.4 minutes per half-life), the amount of a radioactive isotope remaining is found using the formula (1/2)^n, where n is the number of half-lives. Here, n = 60.0 / 20.4 ≈ 3. Therefore, the remaining percentage is (1/2)^3 = 1/8 = 0.125 or 12.5%, which is approximately 13.0%.

Question 10

A freshly prepared sample of curium-243 undergoes 3312 disintegrations per second.After 6.00 yr,the activity of the sample declines to 2755 disintegrations per second.The half-life of curium-243 is ________ yr.

Accepted Answer

The answer of A freshly prepared sample of curium-243 undergoes...

Question 11

The half-life of cobalt-60 is 5.20 yr.How many milligrams of a 2.000-mg sample remain after 6.55 years?

Accepted Answer

The amount of a radioactive substance that remains after a certain period can be calculated using the formula: $A = A_0 \times (1/2)^{t/T}$, where $A$ is the amount remaining, $A_0$ is the initial amount, $t$ is the time elapsed, and $T$ is the half-life of the substance. For cobalt-60, with $A_0 = 2.000$ mg, $t = 6.55$ years, and $T = 5.20$ years, the calculation is $2.000 \times (1/2)^{6.55/5.20} \approx 0.837$ mg.

Question 12

Determine the half-life of a nuclide that loses 38.0% of its mass in 407 hours.

Accepted Answer

The answer of Determine the half-life of a nuclide that...

Question 13

Above what atomic number are there no stable isotopes of any element?

Accepted Answer

Elements with atomic numbers higher than 83 do not have any stable isotopes. This is due to the increasing instability of the nucleus as more and more protons are added, leading to a higher likelihood of radioactive decay.

Question 14

The radioactive decay of ________ is the single greatest source of human exposure to radiation.

Accepted Answer

Radon is a naturally occurring radioactive gas that seeps up from the ground and can accumulate in homes and buildings. It is the leading cause of lung cancer among non-smokers and the second leading cause of lung cancer overall, after smoking. Therefore, it is the single greatest source of human exposure to radiation.

Question 15

Neptunium-239 has a half-life of 2.35 days.How many days must elapse for a sample of ²³⁹Np to decay to 0.100% of its original quantity?

Accepted Answer

To find the number of days for a sample to decay to 0.100% of its original quantity, we use the formula for exponential decay, which involves the half-life of the substance. The formula to calculate the time for a substance to decay to a certain percentage of its original amount is:$$ t = \left( \frac{\log(\text{final amount/original amount})}{\log(0.5)} \right) \times \text{half-life} $$Given that the final amount is 0.100% (or 0.001 in decimal form) of the original amount and the half-life of Neptunium-239 is 2.35 days, we substitute these values into the formula:$$ t = \left( \frac{\log(0.001)}{\log(0.5)} \right) \times 2.35 $$This calculation results in approximately 23.4 days, which means it takes about 23.4 days for a sample of Neptunium-239 to decay to 0.100% of its original quantity.

Question 16

Stable isotopes,with low atomic numbers,have a N/Z ratio of 1.What does that imply?

Accepted Answer

The N/Z ratio of 1 implies that the number of neutrons (N) in the stable isotope equals the number of protons (Z) or atomic number. This is because isotopes are defined by their atomic number (protons), and stable isotopes have an equal number of protons and neutrons to maintain stability.

Question 17

A rock contains 0.313 mg of lead-206 for each milligram of uranium-238.The half-life for the decay of uranium-238 to lead-206 is 4.5 × 10⁹ yr.The rock was formed ________ years ago.

Accepted Answer

The answer of A rock contains 0.313 mg of lead-206...

Question 18

If we start with 1.000 g of cobalt-60,0.675 g will remain after 3.00 yr.This means that the half-life of cobalt-60 is ________ yr.

Accepted Answer

The answer of If we start with 1.000 g of...

Question 19

Complete the following equation of transmutation.  N +  He →  O + ______

Accepted Answer

The answer of Complete the following equation of transmutation. ...

Question 20

Complete the following equation of nuclear fusion.  H +  H →  He + ______

Accepted Answer

The answer of Complete the following equation of nuclear fusion....

Quiz 20: Radioactivity and Nuclear Chemistry

If we start with 1.000 g of strontium-90,0.908 g will remain after 4.00 yr.This means that the half-life of strontium-90 is ________ yr.

The following reaction represents what nuclear process? H + H → He + n