Question 1

Beta decay of ⁶⁰Co produces a beta particle and

Accepted Answer

In beta decay, a neutron in the nucleus is transformed into a proton, an electron (beta particle), and an antineutrino. The atomic number increases by one, but the mass number remains the same. Therefore, Cobalt-60 (Co-60) decays into Nickel-60 (Ni-60).

Question 2

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

Accepted Answer

The half-life of a substance is the time it takes for half of the substance to decay. After one half-life, half of the original amount remains; after two half-lives, half of that amount remains, or one-fourth of the original amount; and so on.In this problem, the half-life of cobalt-60 is 5.20 years. After 6.55 years, the number of half-lives that have passed is 6.55 years / 5.20 years = 1.26.After 1.26 half-lives, the fraction of the original amount that remains is (1/2)^1.26 = 0.418.Therefore, the amount of the original 2.000-mg sample that remains after 6.55 years is 2.000 mg * 0.418 = 0.835 mg.

Question 3

In addition to a beta particle, what is the other product of beta decay of  Pb?

Accepted Answer

Beta decay involves the conversion of a neutron into a proton and an electron (beta particle). The atomic number increases by 1 due to the addition of a proton, while the mass number remains the same. Therefore, the product must have an atomic number of 83 (the next element after lead) and a mass number of 207 (the same as lead). The only option that fits this criteria is Choice C, which represents bismuth.

Question 4

What percentage of a radioactive substance remains after 7.00 half-lives have elapsed?

Accepted Answer

The answer of What percentage of a radioactive substance remains...

Question 5

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 ________ y.

Accepted Answer

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

Question 6

The number of nucleons in a Conucleusis

Accepted Answer

The answer of The number of nucleons in a ...

Question 7

Which nuclide below is most likely to decay by electron capture?

Accepted Answer

The answer of Which nuclide below is most likely to...

Question 8

Which of the following elements would be expected to be particularly stable?

Accepted Answer

The answer of Which of the following elements would be...

Question 9

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

Accepted Answer

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

Question 10

Complete the following equation of nuclear transmutation.  U +  C ?  Cf + 6 ______

Accepted Answer

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

Question 11

Complete the following equation of transmutation.  Al +  He ?  P + ______

Accepted Answer

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

Question 12

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 y. The rock was formed ________ y. ago.

Accepted Answer

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

Question 13

Which of the following nuclides is most likely to undergo beta decay?

Accepted Answer

The answer of Which of the following nuclides is most...

Question 14

Identify the symptom experienced from radiation exposure.

Accepted Answer

Radiation exposure can weaken the immune system, leaving the body more susceptible to illness and infection. Options A, B, C, and D are not symptoms of radiation exposure.

Question 15

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

Accepted Answer

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

Question 16

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 20.4 minutes, half of the sample will remain.
After 40.8 minutes, half of that remaining half will remain (one-fourth of the original sample).
After 61.2 minutes, half of that one-fourth will remain (one-eighth of the original sample).
So, after 60.0 minutes, approximately 13.0% of the sample will remain. This can be calculated using the formula:

% remaining = (1/2)^(t/half-life) x 100%
% remaining = (1/2)^(60/20.4) x 100%
% remaining = 12.959 or approximately 13.0% (rounded to one decimal place)

Question 17

What is the unit for the measure of an absorbed dose of radiation and what is its SI definition?

Accepted Answer

The unit for absorbed dose of radiation is the gray (Gy) which is defined as the deposition of one joule of energy per kilogram of matter. This is the only option that matches both the unit and its definition.

Question 18

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 half-life formula. The formula to calculate the number of half-lives (n) that have passed is given by: n = (log(final amount/original amount)) / log(0.5). For a decay to 0.100% (which is 0.001 of the original amount), we calculate n as follows: n = log(0.001) / log(0.5) ≈ 9.96578 half-lives. Since each half-life of Neptunium-239 is 2.35 days, the total time is about 9.96578 * 2.35 days ≈ 23.4 days.

Question 19

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

Accepted Answer

The percentage of a radioactive sample remaining after a certain time can be calculated using the formula: remaining percentage = (1/2)^(time/half-life) * 100. For strontium-90, after 75.0 years, the calculation is (1/2)^(75.0/28.8) * 100, which equals approximately 16.4%.

Question 20

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

Accepted Answer

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

Quiz 19: Radioactivity and Nuclear Chemistry

Beta decay of 60Co produces a beta particle and