Test Your Science Skills With The Nuclear Physics Quiz – Radioactivity

Initial activity equals lambda times N. Lambda is natural log of two divided by half life. Compute lambda N for each option using proportional reasoning. Option A has the smallest lambda because of the very long half life. Although it has larger moles than C and D, the product lambda N remains smallest due to the factor of half life being twelve times longer than B and far longer than C and D. Short half lives in B, C, and D give much larger lambda even with fewer moles. Therefore A yields the lowest initial decay rate.

Let Nx Ny be numbers at time zero with Nx over Ny equal to 2.50. After time t equals three days, Nx times two to the power minus t over Tx divided by Ny times two to the power minus t over Ty equals 4.20. Rearranging gives two to the power minus t times 1 over Tx minus 1 over Ty equals 4.20 divided by 2.50 which is 1.68. Taking logs gives minus t times natural log two times 1 over Tx minus 1 over Ty equals natural log 1.68. With Ty equal 1.60 days, solve for Tx to get approximately 2.66 days.

Each half value layer halves intensity. After n layers, intensity equals original divided by two to the power n. To reach one sixteenth, we need two to the power n equals sixteen, giving n equals four. Each layer equals two centimeters, so total thickness equals four times two equals eight centimeters. Lesser thicknesses like four or six centimeters correspond to two or three halvings, giving one quarter or one eighth. Greater thickness like ten centimeters exceeds the requirement and would reduce below one sixteenth. Therefore the correct thickness to achieve one sixteenth exactly is eight centimeters.

Activity A equals lambda N with lambda equal natural log two divided by half life. Compute lambda equals 0.693 divided by 5.4 times ten to the power five seconds which is about 1.283 times ten to the power minus six per second. Multiply by N equals 7.2 times ten to the power seventeen to obtain A equals 9.24 times ten to the power eleven becquerel. The closest choice is 9.2 times ten to the power eleven. Options with twelve order and higher are inconsistent because they imply either a significantly shorter half life or many more nuclei than given.

Penetrating power ranks gamma greater than beta greater than alpha in most materials. Gamma radiation is high energy electromagnetic radiation with no charge and negligible mass, interacting mainly through probabilistic processes such as photoelectric effect, Compton scattering, and pair production. These interactions have relatively small cross sections in matter compared with the strong ionization and short range of alpha particles and the moderate interactions of beta particles. Consequently, several centimeters of lead or many centimeters of concrete are required to appreciably attenuate gamma rays, while paper stops alpha and thin metal sheets attenuate beta. Therefore option C is correct.

Exponential decay obeys N equals N0 e to the power minus lambda t. Given N equals 0.70 N0 at t equals twenty minutes. Therefore 0.70 equals e to the power minus lambda times twenty. Taking natural logs gives minus lambda times twenty equals natural log of 0.70 which is about minus 0.356675. Hence lambda equals 0.356675 divided by twenty which equals 0.0178338 per minute. Rounding to four significant figures gives 0.0178 per minute. Alternatives 0.012 and 0.025 are too small or large relative to the measured drop, and 0.035 would predict about fifty percent reduction in twenty minutes.

A straight line in natural log N versus t implies exponential decay N equals N0 e to the power minus lambda t. Intercept natural log N0 equals 3 so N0 equals e to the power 3 which is about 20.085. The slope equals change in natural log N over change in time equals 0 minus 3 over 60 equals minus 0.05. Hence lambda equals 0.05 per second. Substituting gives N equals 20 e to the power minus 0.05 t. Option A matches both the initial value near twenty and the correct decay constant.

Activity decays as A equals A0 times two to the power minus t over T. A fall by ninety percent means A over A0 equals 0.10. Solve 0.10 equals two to the power minus t over T. Take logs to get minus t over T equals log base two of 0.10 which equals minus log base two of 10 which is about minus 3.32193. Therefore t equals 3.32193 times T. With T equal three seconds, t equals about 9.966 seconds. Rounding to reasonable options gives ten seconds, which is option C.

Use N equals N0 times two to the power minus t over T for each nuclide. For X with Tx equal one hour, after six hours Nx equals N0x times two to the power minus six equals N0x divided by sixty four. For Y with Ty equal three hours, after six hours Ny equals N0y times two to the power minus two equals N0y divided by four. Initial ratio N0x equals twelve N0y. Final ratio equals Nx over Ny equals twelve N0y over sixty four divided by N0y over four which simplifies to twelve over sixty four times four equals forty eight over sixty four equals three over four. Option A.

The daughter being an isotope means the atomic number remains the same while the mass number may change. Emitting one alpha reduces atomic number by two and mass number by four. Each beta minus increases atomic number by one with negligible mass change. To keep atomic number unchanged, the total change in atomic number must be zero. One alpha decreases by two. Two beta minus emissions increase by two. Net change equals zero, satisfying the isotope condition. Options with four beta or one beta produce net atomic number changes that do not cancel. Therefore one alpha and two beta is consistent.