301. A fossil sample contains \(\frac{1}{4}\) of the original radioactive carbon activity. The statement most directly supported is:
ⓐ. One half-life has elapsed
ⓑ. Four half-lives have elapsed
ⓒ. Two half-lives have elapsed
ⓓ. No decay has occurred
Correct Answer: Two half-lives have elapsed
Explanation: Activity is proportional to the number of undecayed radioactive nuclei. If the activity is \(\frac{1}{4}\) of the original value, the remaining fraction of radioactive nuclei is also \(\frac{1}{4}\). Since \(\frac{1}{4}=\left(\frac{1}{2}\right)^2\), two half-lives have elapsed. One half-life would leave \(\frac{1}{2}\), not \(\frac{1}{4}\). Four half-lives would leave \(\frac{1}{16}\). The comparison uses fractional activity, not the absolute mass of the fossil.
302. A decay law calculation gives \(N=N_0e^{-0.693}\). This corresponds approximately to:
ⓐ. One half-life
ⓑ. Two half-lives
ⓒ. One mean life
ⓓ. Zero time
Correct Answer: One half-life
Explanation: The decay factor is \(e^{-\lambda t}\). Since \(\ln2=0.693\), the expression \(e^{-0.693}\) is equal to \(e^{-\ln2}\). This is \(\frac{1}{2}\). When \(N=\frac{N_0}{2}\), one half-life has elapsed. One mean life would correspond to \(e^{-1}\), not \(e^{-0.693}\). The number \(0.693\) is therefore the logarithmic marker of half-life in exponential decay.
303. A radioactive sample has a larger decay constant than another isotope. Compared under the same initial number of nuclei, the sample with larger \(\lambda\) has:
ⓐ. Smaller initial activity and longer half-life
ⓑ. Same activity and same half-life
ⓒ. Larger initial activity and shorter half-life
ⓓ. Zero activity because \(\lambda\) is larger
Correct Answer: Larger initial activity and shorter half-life
Explanation: Activity is \(A_c=\lambda N\). If two samples have the same initial number of nuclei \(N_0\), the sample with larger \(\lambda\) has larger initial activity. The half-life is \(T_{1/2}=\frac{\ln2}{\lambda}\), so a larger \(\lambda\) gives a shorter half-life. Thus the isotope with larger decay constant decays faster statistically. This does not mean every individual nucleus decays at the same instant. It means the probability per unit time is higher.
304. A graph of \(\ln\left(\frac{N_0}{N}\right)\) against time \(t\) is plotted for radioactive decay. The slope of the graph is:
ⓐ. \(-\lambda\)
ⓑ. \(\ln N_0\)
ⓒ. \(\lambda\)
ⓓ. \(\frac{1}{\lambda}\)
Correct Answer: \(\lambda\)
Explanation: Starting from the decay law:
\[
N=N_0e^{-\lambda t}
\]
Divide by \(N\) and rearrange:
\[
\frac{N_0}{N}=e^{\lambda t}
\]
Take natural logarithm:
\[
\ln\left(\frac{N_0}{N}\right)=\lambda t
\]
This has the straight-line form \(y=mt\), where the slope is \(\lambda\). The graph has a positive slope because \(\frac{N_0}{N}\) increases as \(N\) decreases. This is different from the \(\ln N\) versus \(t\) graph, whose slope is \(-\lambda\).
305. In a nuclear reaction, the \(Q\)-value is most directly the:
ⓐ. Reaction energy balance
ⓑ. Atomic number of the heaviest product
ⓒ. Radius of the parent nucleus
ⓓ. Number of electrons in the neutral atom only
Correct Answer: Reaction energy balance
Explanation: The \(Q\)-value of a nuclear reaction measures the energy balance of the reaction. If the total mass-energy of the products is less than that of the reactants, energy is released and \(Q\) is positive. If the products have greater total mass-energy, energy must be supplied and \(Q\) is negative. The value is found from the mass difference using \(Q=\Delta mc^2\). It is not an atomic number, radius, or electron count. The sign of \(Q\) tells whether the reaction is exoergic or endoergic.
306. For a reaction written as reactants \(\rightarrow\) products, the \(Q\)-value using masses is:
ⓐ. \(Q=(m_{\text{products}}-m_{\text{reactants}})c^2\)
ⓑ. \(Q=(m_{\text{reactants}}+m_{\text{products}})c^2\)
ⓒ. \(Q=\frac{m_{\text{reactants}}}{m_{\text{products}}}c^2\)
ⓓ. \(Q=(m_{\text{reactants}}-m_{\text{products}})c^2\)
Correct Answer: \(Q=(m_{\text{reactants}}-m_{\text{products}})c^2\)
Explanation: The \(Q\)-value compares total initial mass with total final mass. If reactants have greater mass than products, the missing mass appears as released energy. Therefore, \(Q=(m_{\text{reactants}}-m_{\text{products}})c^2\). A positive result means energy release. Reversing the subtraction would give the opposite sign and would misidentify release as absorption. The formula is a mass-energy balance, so it is not based on adding all masses or forming a mass ratio.
307. A nuclear reaction has total reactant mass \(20.010\,u\) and total product mass \(20.000\,u\). Using \(1\,u\,c^2=931.5\,\text{MeV}\), the \(Q\)-value is:
ⓐ. \(+9.315\,\text{MeV}\)
ⓑ. \(-9.315\,\text{MeV}\)
ⓒ. \(+18.63\,\text{MeV}\)
ⓓ. \(-18.63\,\text{MeV}\)
Correct Answer: \(+9.315\,\text{MeV}\)
Explanation: \( \textbf{Given masses:} \) \(m_{\text{reactants}}=20.010\,u\), and \(m_{\text{products}}=20.000\,u\).
\( \textbf{Mass difference:} \)
\[
\Delta m=m_{\text{reactants}}-m_{\text{products}}
\]
\( \textbf{Substitution:} \)
\[
\Delta m=20.010\,u-20.000\,u=0.010\,u
\]
\( \textbf{Energy conversion:} \)
\[
Q=(0.010)(931.5)\,\text{MeV}
\]
\( \textbf{Calculation:} \)
\[
Q=9.315\,\text{MeV}
\]
\( \textbf{Sign:} \) Since reactant mass is greater than product mass, the reaction releases energy.
\( \textbf{Final answer:} \) The \(Q\)-value is \(+9.315\,\text{MeV}\). The positive sign comes from the decrease in total mass-energy.
308. A proposed reaction has product mass greater than reactant mass. What does this imply about its \(Q\)-value?
ⓐ. \(Q\) is positive and energy is released
ⓑ. \(Q\) is zero for every such reaction
ⓒ. \(Q\) depends only on the number of electrons outside the atom
ⓓ. \(Q\) is negative and energy must be supplied
Correct Answer: \(Q\) is negative and energy must be supplied
Explanation: The \(Q\)-value is \(Q=(m_{\text{reactants}}-m_{\text{products}})c^2\). If the product mass is greater than the reactant mass, the quantity \(m_{\text{reactants}}-m_{\text{products}}\) is negative. Therefore, \(Q\) is negative. A negative \(Q\)-value means the reaction requires energy input. This does not violate conservation of energy; the supplied energy becomes part of the final mass-energy and kinetic energy balance. The sign follows from the mass comparison, not from ordinary electron-shell structure.
309. In the reaction \({}^{2}_{1}H+{}^{3}_{1}H\rightarrow{}^{4}_{2}He+{}^{1}_{0}n+Q\), the reaction is balanced because:
ⓐ. Total \(A\) is \(4\) on both sides and total \(Z\) is \(1\) on both sides
ⓑ. Total \(A\) is \(2\) on both sides and total \(Z\) is \(5\) on both sides
ⓒ. Total \(A\) is \(5\) on both sides and total \(Z\) is \(2\) on both sides
ⓓ. Total \(A\) is not conserved but charge is conserved
Correct Answer: Total \(A\) is \(5\) on both sides and total \(Z\) is \(2\) on both sides
Explanation: Nuclear reactions must conserve total mass number \(A\) and total atomic number \(Z\). On the left side, the mass numbers add to \(2+3=5\), and the atomic numbers add to \(1+1=2\). On the right side, the mass numbers add to \(4+1=5\), and the atomic numbers add to \(2+0=2\). Thus both nucleon number and charge number are conserved. The released energy \(Q\) does not carry mass number or atomic number in this bookkeeping. The reaction is a fusion-type reaction because light nuclei combine to form a heavier nucleus.
310. A reaction \(X+{}^{4}_{2}He\rightarrow{}^{17}_{8}O+{}^{1}_{1}H\) is balanced. The unknown nucleus \(X\) is:
ⓐ. \({}^{14}_{7}N\)
ⓑ. \({}^{13}_{6}C\)
ⓒ. \({}^{16}_{8}O\)
ⓓ. \({}^{12}_{6}C\)
Correct Answer: \({}^{14}_{7}N\)
Explanation: \( \textbf{Use mass-number conservation:} \)
\[
A_X+4=17+1
\]
\[
A_X+4=18
\]
\[
A_X=14
\]
\( \textbf{Use atomic-number conservation:} \)
\[
Z_X+2=8+1
\]
\[
Z_X+2=9
\]
\[
Z_X=7
\]
\( \textbf{Unknown nucleus:} \)
\[
X={}^{14}_{7}N
\]
\( \textbf{Final answer:} \) The missing nucleus is \({}^{14}_{7}N\). Both upper and lower nuclear numbers must be balanced independently.
311. A table gives four possible nuclear reactions. Identify the balanced reaction.
| Row | Reaction |
|---|
| P | \({}^{2}_{1}H+{}^{2}_{1}H\rightarrow{}^{3}_{2}He+{}^{1}_{0}n\) |
| Q | \({}^{2}_{1}H+{}^{2}_{1}H\rightarrow{}^{4}_{2}He+{}^{1}_{0}n\) |
| R | \({}^{4}_{2}He\rightarrow{}^{4}_{3}Li+\gamma\) |
| S | \({}^{14}_{7}N+{}^{4}_{2}He\rightarrow{}^{18}_{10}Ne+{}^{1}_{1}H\) |
ⓐ. Row Q
ⓑ. Row R
ⓒ. Row S
ⓓ. Row P
Correct Answer: Row P
Explanation: Row P is balanced because the left side has total \(A=2+2=4\) and total \(Z=1+1=2\). The right side has \(A=3+1=4\) and \(Z=2+0=2\). Row Q is not balanced in mass number because the right side has \(A=4+1=5\). Row R is not balanced in atomic number because \(2\neq3\). Row S is not balanced in atomic number because the left side has \(Z=9\), while the right side has \(Z=11\). Nuclear equation checking must use both \(A\) and \(Z\), not only a familiar-looking product.
312. Nuclear fission is best described as:
ⓐ. Combination of two light nuclei to form a heavier nucleus
ⓑ. Heavy nucleus splitting
ⓒ. Emission of only a gamma photon with no change in energy
ⓓ. Removal of outer electrons from a neutral atom
Correct Answer: Heavy nucleus splitting
Explanation: Nuclear fission is the splitting of a heavy nucleus into two medium-mass or lighter fragments. The process usually releases energy because the fragments often have a higher binding energy per nucleon than the original heavy nucleus. Neutrons are also commonly emitted in fission, and these can initiate further fissions. Fusion is the combination of light nuclei, not splitting. Ionisation is removal of electrons and is not a nuclear fission process. Fission changes the nucleus itself and must conserve nucleon number and charge number.
313. The fission reaction \({}^{235}_{92}U+{}^{1}_{0}n\rightarrow{}^{141}_{56}Ba+{}^{92}_{36}Kr+3{}^{1}_{0}n+Q\) is checked for mass number. The total mass number on each side is:
ⓐ. \(235\) on both sides
ⓑ. \(233\) on both sides
ⓒ. \(92\) on both sides
ⓓ. \(236\) on both sides
Correct Answer: \(236\) on both sides
Explanation: \( \textbf{Left side mass number:} \)
\[
235+1=236
\]
\( \textbf{Right side mass number:} \)
\[
141+92+3(1)=141+92+3
\]
\[
141+92+3=236
\]
\( \textbf{Mass-number comparison:} \) Both sides have total \(A=236\).
\( \textbf{Charge-number check:} \)
\[
92=56+36+3(0)
\]
\( \textbf{Meaning:} \) The reaction conserves both nucleon number and charge number.
\( \textbf{Final answer:} \) The total mass number is \(236\) on both sides. The extra incoming neutron must be included in the left-side mass number.
314. In a chain reaction, the neutrons released during one fission event are important because they can:
ⓐ. Change gamma rays into alpha particles only
ⓑ. Trigger fission of other suitable nuclei
ⓒ. Remove all electrons from the reactor material
ⓓ. Stop conservation of mass number
Correct Answer: Trigger fission of other suitable nuclei
Explanation: A fission event can release additional neutrons along with energy and fission fragments. These neutrons may be absorbed by other suitable heavy nuclei and cause them to fission. If enough neutrons continue the process, a chain reaction is sustained. Conservation of mass number and charge number still applies to each nuclear reaction. The role of the neutrons is nuclear triggering, not ordinary electron removal. Whether the chain reaction grows, stays steady, or dies out depends on neutron availability and losses.
315. A fission chain reaction becomes self-sustaining when:
ⓐ. Every emitted neutron must escape the material
ⓑ. The material contains no nuclei
ⓒ. Gamma rays replace all neutrons in the process
ⓓ. One neutron from each fission causes another fission on average
Correct Answer: One neutron from each fission causes another fission on average
Explanation: A self-sustaining chain reaction needs enough neutrons to continue the fission sequence. If, on average, one neutron from each fission causes another fission, the reaction can remain steady. If fewer than one effective neutron continues the process, the chain reaction dies out. If more than one effective neutron continues the process, the reaction grows. Neutrons can be lost by escape or absorption without fission, so their effective number matters. The condition is statistical and applies to a large number of fission events.
316. A block of fissile material is too small to maintain a chain reaction mainly because:
ⓐ. Its protons have no charge
ⓑ. Excess neutron leakage
ⓒ. Binding energy per nucleon becomes exactly zero
ⓓ. The mass number of every nucleus becomes \(1\)
Correct Answer: Excess neutron leakage
Explanation: In a small piece of fissile material, many neutrons produced by fission can escape through the surface. Escaped neutrons cannot trigger additional fissions inside the material. If the average number of effective neutrons falls below the required value, the chain reaction cannot sustain itself. Increasing size reduces the relative surface loss of neutrons. This idea is connected with critical size or critical mass. It is not caused by protons losing charge or nuclei losing all binding energy.
317. In a controlled nuclear reactor, the purpose of control rods is to:
ⓐ. Increase the number of protons in every nucleus
ⓑ. Absorb excess neutrons
ⓒ. Convert all gamma rays into chemical bonds
ⓓ. Increase radioactive decay constant of fuel instantly
Correct Answer: Absorb excess neutrons
Explanation: Control rods are made of materials that absorb neutrons effectively. By absorbing some of the neutrons, they reduce the number available to cause further fissions. This helps regulate the chain reaction and keep it controlled. If more control rod material is inserted, neutron absorption increases and the reaction rate can decrease. If rods are withdrawn, more neutrons remain available for fission. Control rods do not change the atomic number of every fuel nucleus; they control neutron population.
318. The moderator in a thermal nuclear reactor is used mainly to:
ⓐ. Absorb all neutrons permanently
ⓑ. Increase the charge of alpha particles
ⓒ. Slow fast neutrons
ⓓ. Remove electrons from uranium atoms
Correct Answer: Slow fast neutrons
Explanation: Fission of suitable fuel such as \({}^{235}_{92}U\) is more effectively caused by slow neutrons. Neutrons produced in fission are often fast. A moderator reduces their kinetic energy through collisions without absorbing too many of them. Slow or thermal neutrons can then induce further fissions more efficiently. Control rods, not moderators, are mainly used to absorb excess neutrons. The moderator changes neutron speed, not the charge of alpha particles or the electron count of atoms.
319. The coolant in a nuclear reactor is used to:
ⓐ. Carry heat away from the reactor core
ⓑ. Balance the atomic number in fission equations
ⓒ. Convert neutrons into protons in the fuel
ⓓ. Stop all nuclear reactions by chemical bonding
Correct Answer: Carry heat away from the reactor core
Explanation: Nuclear fission releases energy mainly as kinetic energy of fragments and other particles, which becomes heat in the reactor core. The coolant carries this heat away from the core. The heat can then be used to produce steam and drive turbines in a power plant. The coolant is different from the moderator and control rods, although in some designs one material may have more than one role. It does not balance nuclear equations or change neutron number directly. Its main function is thermal energy transfer.
320. Match the reactor component with its main function.
| Column I | Column II |
|---|
| P. Fuel | 1. Undergoes fission and releases energy |
| Q. Moderator | 2. Slows down fast neutrons |
| R. Control rods | 3. Absorb excess neutrons |
| S. Coolant | 4. Carries heat away from the core |
ⓐ. P-2, Q-1, R-3, S-4
ⓑ. P-1, Q-2, R-3, S-4
ⓒ. P-1, Q-3, R-2, S-4
ⓓ. P-4, Q-2, R-1, S-3
Correct Answer: P-1, Q-2, R-3, S-4
Explanation: The fuel contains fissile nuclei that undergo fission and release energy. The moderator slows down fast neutrons so that they can more effectively cause further fission in suitable fuel. Control rods absorb excess neutrons and regulate the chain reaction. The coolant removes heat from the reactor core and transfers it for useful energy conversion. These functions are related but not interchangeable. Confusing moderator and control rods is especially common because both affect neutrons, but one slows them while the other absorbs them.
