Call Now to Set Up Tutoring:

(888) 888-0446

AP Physics 2 : Quantum and Nuclear Physics

Study concepts, example questions & explanations for AP Physics 2

Create An Account

All AP Physics 2 Resources

6 Diagnostic Tests 149 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #21 : Atomic And Nuclear Physics

A scientist takes a reading of a radioactive material, which has an activity of 1570 Bq . 15 minutes later, it has an activity of 925 Bq .

Determine the activity at 25 minutes after the initial reading.

Possible Answers:

A=677Bq

None of these

A=899Bq

A=150Bq

A=650Bq

Correct answer:

A=650Bq

Explanation:

Use the relationship:

$A=A_0e^{-\lambda t}$

Here, is the activity at a given time, A_0 is the initial activity, $\lambda$ is the radioactive decay constant, and is the time passed since the initial reading.

Rearranging the equation to solve for $\lambda$ .

$\frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$

Convert minutes to seconds and plug in values.

$\frac{(-ln(\frac{925}{1570}))}{900s}=\lambda$

$\frac{5.88*10^{-4}}{s}=\lambda$

Again use the relationship:

$A=A_0e^{-\lambda t}$

Using the new , which is equal to 25min=1500s

$A=1570e^{-5.88*10^{-4} 1500}$

A=650Bq

Report an Error

Example Question #371 : Ap Physics 2

A scientist takes a reading of a radioactive material, which has an activity of 1570 Bq . 15 minutes later, it has an activity of 925 Bq .

Determine the half-life of this isotope.

Possible Answers:

None of these

$395s=t_{\frac{1}{2}}$

$1007s=t_{\frac{1}{2}}$

$1395s=t_{\frac{1}{2}}$

$1180s=t_{\frac{1}{2}}$

Correct answer:

$1180s=t_{\frac{1}{2}}$

Explanation:

Use the relationship:

$A=A_0e^{-\lambda t}$

Here, is the activity at a given time, A_0 is the initial activity, $\lambda$ is the radioactive decay constant, and is the time passed since the initial reading.

Rearrange the equation to solve for $\lambda$ .

$\frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$

Convert minutes to seconds and pluge in values.

$\frac{(-ln(\frac{925}{1570}))}{900s}=\lambda$

$\frac{5.88*10^{-4}}{s}=\lambda$

Use the relationship:

$\frac{ln(2))}{\lambda}=t_{1/2}$

Plug in the calculated value for $\lambda$ and solve

$\frac{ln(2))}{5.88*10^{-4}}=t_{\frac{1}{2}}$

$1180s=t_{\frac{1}{2}}$

Report an Error

Example Question #13 : Radioactive Nuclear Decay

A scientist takes a reading of a radioactive material, which has an activity of 1570 Bq . 15 minutes later, it has an activity of 925 Bq .

Determine the nuclear decay constant of this isotope.

Possible Answers:

$\frac{5.88*10^{-4}}{s}=\lambda$

None of these

$\frac{3.98*10^{-4}}{s}=\lambda$

$\frac{8.55*10^{-4}}{s}=\lambda$

$\frac{1.13*10^{-4}}{s}=\lambda$

Correct answer:

$\frac{5.88*10^{-4}}{s}=\lambda$

Explanation:

Use the relationship:

$A=A_0e^{-\lambda t}$

Here, is the activity at a given time, A_0 is the initial activity, $\lambda$ is the radioactive decay constant, and is the time passed since the initial reading.

Rearrange the equation to solve for $\lambda$ .

$\frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$

Convert minutes to seconds and plug in values.

$\frac{(-ln(\frac{925}{1570}))}{900s}=\lambda$

$\frac{5.88*10^{-4}}{s}=\lambda$

Report an Error

Example Question #14 : Radioactive Nuclear Decay

A test is done on a sample of a newly discovered radioactive nuclei, which has an activity of 1990 Bq . 10 minutes later, it has an activity of 220 Bq .

Determine the half life of this nuclei.

Possible Answers:

282seconds

181seconds

181minutes

None of these

113seconds

Correct answer:

181seconds

Explanation:

Using the relationship:

$A=A_0e^{-\lambda t}$

Here, is the activity at a given time, A_0 is the intial activity, $\lambda$ is the radioactive decay constant, and is the time passed since the initial reading.

Rearranging the equation to solve for $\lambda$ .

$\frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$

Converting minutes to seconds and plugging in values.

$\frac{(-ln(\frac{200}{1990}))}{600s}=\lambda$

$\frac{3.83*10^{-3}}{s}=\lambda$

Using the relationship

$\frac{ln(2))}{\lambda}=t_{1/2}$

Plugging in the calculated value for $\lambda$

$\frac{ln(2))}{3.83*10^{-3}}=t_{1/2}$

$181s=t_{1/2}$

Report an Error

Example Question #41 : Quantum And Nuclear Physics

A scientist discovers a new radioactive nuclei. She runs a test on a sample and finds it has an activity of 3.3*10^4Bq . 17 years later, it has an activity of 2.9*10^4 Bq .

Determine the decay constant.

Possible Answers:

$\frac{7.6*10^{-3}}{year}=\lambda$

$\frac{3.3*10^{-1}}{year}=\lambda$

$\frac{2.2*10^{-5}}{year}=\lambda$

$\frac{9.9*10^{-2}}{year}=\lambda$

None of these

Correct answer:

$\frac{7.6*10^{-3}}{year}=\lambda$

Explanation:

Using the relationship:

$A=A_0e^{-\lambda t}$

Here, is the activity at a given time, A_0 is the intial activity, $\lambda$ is the radioactive decay constant, and is the time passed since the initial reading.

Rearranging the equation to solve for $\lambda$ .

$\frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$

Converting minutes to seconds and plugging in values.

$\frac{(-ln(\frac{2.9*10^4 Bq}{3.3*10^4 Bq}))}{17years}=\lambda$

$\frac{7.6*10^{-3}}{year}=\lambda$

Report an Error

Example Question #42 : Quantum And Nuclear Physics

A scientist discovers a new radioactive nuclei. She runs a test on a sample and finds it has an activity of 3.3*10^4Bq . 17 years later, it has an activity of 2.9*10^4 Bq .

Determine the half life.

Possible Answers:

$31.1years=t_{1/2}$

$17.2years=t_{1/2}$

$91.2years=t_{1/2}$

$22.2years=t_{1/2}$

None of these

Correct answer:

$91.2years=t_{1/2}$

Explanation:

Using the relationship:

$A=A_0e^{-\lambda t}$

Here, is the activity at a given time, A_0 is the intial activity, $\lambda$ is the radioactive decay constant, and is the time passed since the initial reading.

Rearranging the equation to solve for $\lambda$ .

$\frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$

Converting minutes to seconds and plugging in values.

$\frac{(-ln(\frac{2.9*10^4 Bq}{3.3*10^4 Bq}))}{17years}=\lambda$

$\frac{7.6*10^{-3}}{year}=\lambda$

Using the relationship

$\frac{ln(2))}{\lambda}=t_{1/2}$

Plugging in the calculated value for $\lambda$

$\frac{ln(2))}{7.6*10^{-3}}=t_{1/2}$

$91.2years=t_{1/2}$

Report an Error

Example Question #23 : Atomic And Nuclear Physics

A scientist discovers a new radioactive nuclei. She runs a test on a sample and finds it has an activity of 3.3*10^4Bq . 17 years later, it has an activity of 2.9*10^4 Bq .

Determine the number of radioactive atoms in the initial sample.

Possible Answers:

None of these

$N=3.33*10^{12}$

$N=5.71*10^{6}$

$N=5.71*10^{12}$

$N=2.75*10^{12}$

Correct answer:

$N=5.71*10^{12}$

Explanation:

Using the relationship:

$A=A_0e^{-\lambda t}$

Here, is the activity at a given time, A_0 is the intial activity, $\lambda$ is the radioactive decay constant, and is the time passed since the initial reading.

Rearranging the equation to solve for $\lambda$ .

$\frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$

Converting minutes to seconds and plugging in values.

$\frac{(-ln(\frac{2.9*10^4 Bq}{3.3*10^4 Bq}))}{17years}=\lambda$

$\frac{7.6*10^{-3}}{year}=\lambda$

It is then necessary to use the relationship

$A= \lambda N$

Changing the units of the decay constant to be consistent with the activity given.

$\frac{7.6*10^{-3}}{year}*\frac{1year}{365.25days}*\frac{1day}{3600seconds}=\frac{5.78*10^{-9}}{second}$

Using the initial activity and the calculated decay constant:

$3.3*10^{4}= \frac{5.78*10^{-9}}{s} N$

$N=5.71*10^{12}$

Report an Error

Example Question #18 : Radioactive Nuclear Decay

A scientist discovers a new radioactive nuclei. She runs a test on a sample and finds it has an activity of 3.3*10^4Bq . 17 years later, it has an activity of 2.9*10^4 Bq .

Determine the activity 23 years after the original reading.

Possible Answers:

A=3.88*10^4Bq

A=2.77*10^4Bq

A=1.22*10^4Bq

None of these

There will be no activity left

Correct answer:

A=2.77*10^4Bq

Explanation:

Using the relationship:

$A=A_0e^{-\lambda t}$

Here, is the activity at a given time, A_0 is the intial activity, $\lambda$ is the radioactive decay constant, and is the time passed since the initial reading.

Rearranging the equation to solve for $\lambda$ .

$\frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$

Converting minutes to seconds and plugging in values.

$\frac{(-ln(\frac{2.9*10^4 Bq}{3.3*10^4 Bq}))}{17years}=\lambda$

$\frac{7.6*10^{-3}}{year}=\lambda$

Again using the relationship

$A=A_0e^{-\lambda t}$

Using the new

$A=3.3*10^{4}e^{-{7.6*10^{-3} 23}}$

A=2.77*10^4Bq

Report an Error

Example Question #19 : Radioactive Nuclear Decay

A test is done on a sample of a newly discovered radioactive nuclei, which has an activity of 1990 Bq . 10 minutes later, it has an activity of 220 Bq .

Determine the activity 12 minutes after the initial reading.

Possible Answers:

330bq

None of these

83.1bq

31.8bq

75bq

Correct answer:

31.8bq

Explanation:

Using the relationship:

$A=A_0e^{-\lambda t}$

Here, is the activity at a given time, A_0 is the intial activity, $\lambda$ is the radioactive decay constant, and is the time passed since the initial reading.

Rearranging the equation to solve for $\lambda$ .

$\frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$

Converting minutes to seconds and plugging in values.

$\frac{(-ln(\frac{200}{1990}))}{600s}=\lambda$

$\frac{3.83*10^{-3}}{s}=\lambda$

Again using the relationship

$A=A_0e^{-\lambda t}$

Using the new , which is equal to 12min=720seconds

$A=1990e^{-3.83*10^{-3} 1080}$

A=31.8bq

Report an Error

Example Question #20 : Radioactive Nuclear Decay

A scientist takes a sample of a newly discovered radioactive element, which has an activity of 1990 Bq . 10 minutes later, it has an activity of 220 Bq .

Determine the number of radioactive nuclei in the original sample.

Possible Answers:

N=1.44*10^5

N=3.77*10^5

None of these

N=9.85*10^5

N=5.42*10^5

Correct answer:

N=5.42*10^5

Explanation:

Use the relationship:

$A=A_0e^{-\lambda t}$

Where is the activity at a given time, A_0 is the initial activity, $\lambda$ is the radioactive decay constant and is the time passed since the initial reading.

Rearrange to solve for $\lambda$ .

$\frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$

Convert minutes to seconds and plug in values.

$\frac{(-ln(\frac{220 Bq}{1990 Bq}))}{600s}=\lambda$

$\frac{3.67*10^{-3}}{s}=\lambda$

It is then necessary to use the relationship:

$A= \lambda N$

Use the initial activity and the calculated decay constant:

$1990= \frac{3.67*10^{-3}}{s} N$

N=5.42*10^5

Report an Error

View Tutors

Rebecca
Certified Tutor

University of South Florida, Bachelor of Science, Special Education. University of New England, Master of Science, Education.

View Tutors

Abhya
Certified Tutor

O.P JINDAL GLOBAL LAW SCHOOL, Master's/Graduate, MASTERS IN LAW.

View Tutors

Usha
Certified Tutor

SRM University, Bachelor of Technology, Biotechnology. University of Ottawa, Master of Science, Biochemistry.

All AP Physics 2 Resources

6 Diagnostic Tests 149 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

SSAT Tutors in Houston, Statistics Tutors in Seattle, Computer Science Tutors in Seattle, Spanish Tutors in Miami, MCAT Tutors in Chicago, Physics Tutors in Atlanta, MCAT Tutors in Philadelphia, French Tutors in New York City, SAT Tutors in San Francisco-Bay Area, Math Tutors in San Diego

Popular Courses & Classes

GMAT Courses & Classes in San Francisco-Bay Area, LSAT Courses & Classes in Boston, MCAT Courses & Classes in New York City, ACT Courses & Classes in San Francisco-Bay Area, ISEE Courses & Classes in Houston, SAT Courses & Classes in San Diego, ACT Courses & Classes in Atlanta, GRE Courses & Classes in San Diego, SSAT Courses & Classes in Denver, MCAT Courses & Classes in Dallas Fort Worth

Popular Test Prep

LSAT Test Prep in Phoenix, ISEE Test Prep in Miami, LSAT Test Prep in Boston, MCAT Test Prep in Boston, ISEE Test Prep in Phoenix, ACT Test Prep in San Francisco-Bay Area, GMAT Test Prep in San Francisco-Bay Area, SSAT Test Prep in San Francisco-Bay Area, SAT Test Prep in San Diego, SSAT Test Prep in Denver

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

AP Physics 2 : Quantum and Nuclear Physics

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #21 : Atomic And Nuclear Physics

Example Question #371 : Ap Physics 2

Example Question #13 : Radioactive Nuclear Decay

Example Question #14 : Radioactive Nuclear Decay

Example Question #41 : Quantum And Nuclear Physics

Example Question #42 : Quantum And Nuclear Physics

Example Question #23 : Atomic And Nuclear Physics

Example Question #18 : Radioactive Nuclear Decay

Example Question #19 : Radioactive Nuclear Decay

Example Question #20 : Radioactive Nuclear Decay

All AP Physics 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: