Nuclear Chemistry — Hard Practice Quiz
A Chemistry cheat sheet for Nuclear Chemistry — every key formula with its symbols defined — plus a hard-level practice quiz to test recall.
Formulas & key concepts
First-Order Rate Law for Nuclear Decay: Relates the amount of radioactive substance remaining to time.
Where: \(N_t\) = amount at time t, \(N_0\) = initial amount, \(k\) = decay constant, \(t\) = time
Decay Constant and Half-Life: Relates decay constant to half-life of radioactive isotope.
Where: \(t_{1/2}\) = half-life
Einstein's Mass-Energy Equation: Relates mass to energy in nuclear reactions.
Where: \(m\) = mass, \(c\) = speed of light
Practice quiz
An isotope has a half-life of $T$. If an initial sample of $N_0$ atoms decays for a time $t = 2T$, what fraction of the original mass remains?
- $1/2$
- $1/4$
- $1/e$
- $e^{-2}$
Answer: $1/4$
If a radioactive sample decays to $1/e$ of its initial amount in time $t$, what is its half-life in terms of $t$?
- $t/\ln(2)$
- $t \ln(2)$
- $t/0.693$
- $t/e$
Answer: $t \ln(2)$
Isotope A has a half-life of $T_A$. Isotope B has a half-life of $T_B = 2T_A$. If both start with the same number of atoms, after what time will the ratio of remaining atoms $N_{t,B}/N_{t,A}$ be $e$?
- $T_A/\ln(2)$
- $2T_A/\ln(2)$
- $T_A \ln(2)$
- $2T_A \ln(2)$
Answer: $2T_A/\ln(2)$
A sample of a radioactive substance initially contains $X$ grams. After $Y$ years, $Z$ grams remain. What is the decay constant $k$ for this substance?
- $\frac{1}{Y} \ln(\frac{Z}{X})$
- $Y \ln(\frac{X}{Z})$
- $\frac{1}{Y} \ln(\frac{X}{Z})$
- $\frac{0.693}{Y} \ln(\frac{X}{Z})$
Answer: $\frac{1}{Y} \ln(\frac{X}{Z})$
Consider two radioactive isotopes, X and Y, with half-lives $t_{1/2,X}$ and $t_{1/2,Y}$ respectively. If both isotopes undergo a decay process that converts the same mass defect $\Delta m$ per atom into energy, and we start with equal initial masses of X and Y, which isotope will release energy at a faster initial rate?
- Isotope X, if $t_{1/2,X} > t_{1/2,Y}$
- Isotope Y, if $t_{1/2,X} > t_{1/2,Y}$
- Both release energy at the same initial rate.
- It depends on the specific value of $\Delta m$.
Answer: Isotope Y, if $t_{1/2,X} > t_{1/2,Y}$
A radioactive sample's activity drops to $1/8$ of its initial value in $12$ hours. What is the decay constant $k$ in units of $\text{hr}^{-1}$?
- $\frac{\ln(2)}{12} \text{ hr}^{-1}$
- $\frac{\ln(8)}{12} \text{ hr}^{-1}$
- $\frac{0.693}{8} \text{ hr}^{-1}$
- $\frac{1}{12} \text{ hr}^{-1}$
Answer: $\frac{\ln(8)}{12} \text{ hr}^{-1}$
An ancient artifact is found to have $1/16$ of the original amount of Carbon-14. If the half-life of Carbon-14 is $5730$ years, how many years old is the artifact?
- $5730 \text{ years}$
- $11460 \text{ years}$
- $22920 \text{ years}$
- $28650 \text{ years}$
Answer: $22920 \text{ years}$
Isotope A has a decay constant $k_A$. Isotope B has a decay constant $k_B = 2k_A$. If both start with the same initial number of atoms, after how many half-lives of Isotope A will Isotope B have decayed to $1/4$ of its initial amount?
- $0.5$
- $1$
- $2$
- $4$
Answer: $1$
If a nuclear reaction releases energy $E$, and the speed of light is $c$, what is the equivalent mass defect $\Delta m$ in terms of $E$ and $c$? If $c$ were twice its actual value, how would the mass defect required to release the same energy $E$ change?
- $\Delta m = E/c^2$; it would be $1/2$ of the original.
- $\Delta m = E/c^2$; it would be $1/4$ of the original.
- $\Delta m = Ec^2$; it would be $4$ times the original.
- $\Delta m = Ec^2$; it would be $2$ times the original.
Answer: $\Delta m = E/c^2$; it would be $1/4$ of the original.
A radioactive isotope has a half-life of $10 \text{ years}$. Each decay event results in a mass defect of $1.5 \times 10^{-28} \text{ kg}$. If an initial sample contains $N_0 = 1.0 \times 10^{24}$ atoms, what is the total energy released from the sample after $20 \text{ years}$? (Use $c = 3.0 \times 10^8 \text{ m/s}$)
- $2.53 \times 10^{12} \text{ J}$
- $5.06 \times 10^{12} \text{ J}$
- $1.01 \times 10^{13} \text{ J}$
- $1.52 \times 10^{13} \text{ J}$
Answer: $1.01 \times 10^{13} \text{ J}$
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