101. A statement says, “Because the nucleus is extremely dense, it must occupy most of the atom's volume.” The best response is:
ⓐ. The statement is valid because high density always means large volume
ⓑ. The statement is valid only for light atoms
ⓒ. Not valid; the nucleus is dense but tiny
ⓓ. The statement is not valid because nuclei have no mass
Correct Answer: Not valid; the nucleus is dense but tiny
Explanation: Density is mass per unit volume, so a very large density can occur when a large amount of mass is concentrated in a tiny volume. The nucleus contains almost all the mass of the atom, but its radius is much smaller than the atomic radius. Therefore, it occupies only a very small fraction of the atom's volume. The high density of nuclear matter does not imply that the nucleus fills most of the atom. Most of the atom's volume is associated with the electron region. The statement confuses density with size.
102. In a simplified scattering picture, most alpha particles pass almost straight through a thin foil, while a few are deflected through large angles. This pattern suggests that:
ⓐ. Positive charge is uniformly spread throughout the atom
ⓑ. Small central region of concentrated charge and mass
ⓒ. Electrons carry most of the atomic mass
ⓓ. The nucleus has the same size as the atom
Correct Answer: Small central region of concentrated charge and mass
Explanation: If positive charge were spread uniformly throughout the atom, large-angle deflections of alpha particles would not be expected from a tiny central region. The fact that most particles pass through shows that most of the atom is empty space. The rare large deflections show that a strong repulsive centre exists in a very small region. Since alpha particles are positively charged, this centre must be positively charged. The concentration of mass in the same region explains why the alpha particle can be deflected strongly. This scattering pattern supports the nuclear model of the atom.
103. The mass defect of a stable bound nucleus refers to the difference between:
ⓐ. The atomic mass and the electron mass only
ⓑ. The number of protons and the number of neutrons
ⓒ. The nuclear radius and the atomic radius
ⓓ. Separated-nucleon mass minus actual nuclear mass
Correct Answer: Separated-nucleon mass minus actual nuclear mass
Explanation: Mass defect is a mass difference, not a difference in particle number or size. For a nucleus made of \(Z\) protons and \(N\) neutrons, the sum of the masses of the separated nucleons is slightly greater than the actual mass of the bound nucleus. This missing mass is called the mass defect \(\Delta m\). It is connected with the energy released when the nucleus is formed from separated nucleons. The same energy must be supplied to separate the nucleus completely into free protons and neutrons. A stable bound nucleus has positive mass defect because the bound system has lower mass-energy than its separated parts.
104. A nucleus contains \(Z\) protons and \(N\) neutrons. If \(M_{\text{nucleus}}\) is its actual nuclear mass, the mass defect is written as:
ⓐ. \(\Delta m=M_{\text{nucleus}}-Zm_p-Nm_n\)
ⓑ. \(\Delta m=Zm_e+Nm_n-M_{\text{nucleus}}\)
ⓒ. \(\Delta m=Zm_p+Nm_n-M_{\text{nucleus}}\)
ⓓ. \(\Delta m=A-M_{\text{nucleus}}\)
Correct Answer: \(\Delta m=Zm_p+Nm_n-M_{\text{nucleus}}\)
Explanation: The mass defect compares the mass of free separated nucleons with the mass of the bound nucleus. A nucleus with \(Z\) protons and \(N\) neutrons has separated-nucleon mass \(Zm_p+Nm_n\). The actual bound nuclear mass is \(M_{\text{nucleus}}\). Therefore, the mass defect is \(\Delta m=Zm_p+Nm_n-M_{\text{nucleus}}\). Electrons are not included in this nuclear-mass formula because \(M_{\text{nucleus}}\) is the mass of the bare nucleus. Writing the expression in the reverse order would make \(\Delta m\) negative for an ordinary stable bound nucleus.
105. A statement says, “The actual mass of a nucleus is equal to the exact sum of the masses of its free protons and neutrons.” The best evaluation is:
ⓐ. It is valid because nucleons do not interact inside a nucleus
ⓑ. Not valid; binding lowers nuclear mass-energy
ⓒ. It is valid only for heavy nuclei
ⓓ. It is not valid because a nucleus contains only electrons
Correct Answer: Not valid; binding lowers nuclear mass-energy
Explanation: When separated protons and neutrons form a bound nucleus, energy is released. By mass-energy equivalence, this released energy corresponds to a decrease in mass. Therefore, the actual mass of the bound nucleus is less than the sum of the masses of its separated nucleons. This difference is the mass defect. The nucleons interact strongly inside the nucleus, so they cannot be treated as if binding had no energy effect. The mass difference is small on the \(u\) scale but important on the \(\text{MeV}\) energy scale.
106. A nuclear data line gives \(Z=3\), \(N=4\), \(m_p=1.0073\,u\), \(m_n=1.0087\,u\), and \(M_{\text{nucleus}}=7.0144\,u\). The mass defect is closest to:
ⓐ. \(0.0423\,u\)
ⓑ. \(0.0023\,u\)
ⓒ. \(0.2330\,u\)
ⓓ. \(7.0144\,u\)
Correct Answer: \(0.0423\,u\)
Explanation: \( \textbf{Given:} \) \(Z=3\), \(N=4\), \(m_p=1.0073\,u\), \(m_n=1.0087\,u\), and \(M_{\text{nucleus}}=7.0144\,u\).
\( \textbf{Required quantity:} \) Mass defect \(\Delta m\).
\( \textbf{Mass defect relation:} \)
\[
\Delta m=Zm_p+Nm_n-M_{\text{nucleus}}
\]
\( \textbf{Separated proton mass:} \)
\[
Zm_p=3(1.0073)=3.0219\,u
\]
\( \textbf{Separated neutron mass:} \)
\[
Nm_n=4(1.0087)=4.0348\,u
\]
\( \textbf{Total separated nucleon mass:} \)
\[
3.0219\,u+4.0348\,u=7.0567\,u
\]
\( \textbf{Mass defect:} \)
\[
\Delta m=7.0567\,u-7.0144\,u=0.0423\,u
\]
\( \textbf{Final answer:} \) The mass defect is \(0.0423\,u\). The actual nucleus has a smaller mass than the separated nucleons, and the difference is the nuclear mass defect.
107. For a stable bound nucleus, the sign of mass defect \(\Delta m\) is usually:
ⓐ. Negative, because the bound nucleus has greater mass than separated nucleons
ⓑ. Zero, because mass is always exactly conserved without energy change
ⓒ. Alternating, because protons are positive and neutrons are neutral
ⓓ. Positive; free nucleons have greater mass
Correct Answer: Positive; free nucleons have greater mass
Explanation: A stable bound nucleus has lower mass-energy than the same nucleons when they are completely separated. The difference between the separated-nucleon mass and the actual nuclear mass is the mass defect. With the convention \(\Delta m=Zm_p+Nm_n-M_{\text{nucleus}}\), the value is positive for an ordinary bound nucleus. This positive mass defect corresponds to binding energy through \(B=\Delta mc^2\). It does not mean mass has disappeared without trace; it means the equivalent energy has been released during formation. The sign depends on subtracting the bound nuclear mass from the free-nucleon mass sum.
108. Using nuclear data \(Z=2\), \(N=2\), \(m_p=1.0073\,u\), \(m_n=1.0087\,u\), and \(M_{\text{nucleus}}=4.0015\,u\), the mass defect is:
ⓐ. \(0.0152\,u\)
ⓑ. \(0.0609\,u\)
ⓒ. \(4.0319\,u\)
ⓓ. \(0.0305\,u\)
Correct Answer: \(0.0305\,u\)
Explanation: \( \textbf{Given data:} \) \(Z=2\), \(N=2\), \(m_p=1.0073\,u\), \(m_n=1.0087\,u\), and \(M_{\text{nucleus}}=4.0015\,u\).
\( \textbf{Required quantity:} \) Mass defect \(\Delta m\).
\( \textbf{Formula:} \)
\[
\Delta m=Zm_p+Nm_n-M_{\text{nucleus}}
\]
\( \textbf{Mass of separated protons:} \)
\[
Zm_p=2(1.0073)=2.0146\,u
\]
\( \textbf{Mass of separated neutrons:} \)
\[
Nm_n=2(1.0087)=2.0174\,u
\]
\( \textbf{Total separated-nucleon mass:} \)
\[
2.0146\,u+2.0174\,u=4.0320\,u
\]
\( \textbf{Mass defect:} \)
\[
\Delta m=4.0320\,u-4.0015\,u=0.0305\,u
\]
\( \textbf{Final answer:} \) The mass defect is \(0.0305\,u\). The actual nucleus has smaller mass than the separated nucleons because the bound state has lower mass-energy.
109. A nucleus is assembled from free nucleons and energy is released. The released energy is related to:
ⓐ. The mass defect of the nucleus
ⓑ. The increase in atomic radius
ⓒ. The number of orbital electrons only
ⓓ. The chemical symbol only
Correct Answer: The mass defect of the nucleus
Explanation: When free protons and neutrons combine to form a bound nucleus, the final bound nucleus has less mass than the initial separated nucleons. The mass difference is the mass defect \(\Delta m\). According to mass-energy equivalence, the released energy is \(\Delta mc^2\). This energy is called the binding energy when viewed as the energy needed to separate the nucleus again. It is a nuclear energy effect, not an ordinary atomic-shell or chemical-symbol effect. The chemical symbol identifies the element, but the energy comes from the change in nuclear mass-energy.
110. The quantity called binding energy of a nucleus is best defined as the energy required to:
ⓐ. Remove all electrons from the neutral atom only
ⓑ. Separate the nucleus into free nucleons
ⓒ. Convert every proton into a neutron without energy change
ⓓ. Increase the nuclear radius to the atomic radius
Correct Answer: Separate the nucleus into free nucleons
Explanation: Binding energy is the energy needed to break a nucleus into its separated nucleons. Those separated nucleons are free protons and free neutrons, not atomic electrons. The same amount of energy is released when the nucleus is formed from those separated nucleons. This two-way meaning follows from conservation of energy and mass-energy equivalence. Removing electrons from an atom is ionisation, which is an atomic process, not nuclear binding. Binding energy measures how strongly the nucleons are held together inside the nucleus.
111. If a nucleus has mass defect \(\Delta m\), its binding energy \(B\) is:
ⓐ. \(B=\Delta mc^2\)
ⓑ. \(B=\frac{\Delta m}{c^2}\)
ⓒ. \(B=\Delta m+c^2\)
ⓓ. \(B=\frac{c^2}{\Delta m}\)
Correct Answer: \(B=\Delta mc^2\)
Explanation: Binding energy is the energy equivalent of the mass defect. Einstein's mass-energy relation connects a mass difference with energy through multiplication by \(c^2\). Therefore, \(B=\Delta mc^2\). If \(\Delta m\) is in \(\text{kg}\), the energy comes out in \(\text{J}\). If \(\Delta m\) is in \(u\), the conversion \(1\,u\,c^2\approx931.5\,\text{MeV}\) is convenient. Dividing by \(c^2\) would convert energy into mass, not mass defect into binding energy.
112. For \(\Delta m=0.0305\,u\), using \(1\,u\,c^2=931.5\,\text{MeV}\), the nuclear binding energy is closest to:
ⓐ. \(14.2\,\text{MeV}\)
ⓑ. \(30.5\,\text{MeV}\)
ⓒ. \(28.4\,\text{MeV}\)
ⓓ. \(56.8\,\text{MeV}\)
Correct Answer: \(28.4\,\text{MeV}\)
Explanation: \( \textbf{Given:} \) \(\Delta m=0.0305\,u\).
\( \textbf{Conversion factor:} \) \(1\,u\,c^2=931.5\,\text{MeV}\).
\( \textbf{Required quantity:} \) Binding energy \(B\).
\( \textbf{Binding energy relation:} \)
\[
B=\Delta mc^2
\]
\( \textbf{Using the \(u\)-to-\(\text{MeV}\) conversion:} \)
\[
B=(0.0305)(931.5)\,\text{MeV}
\]
\( \textbf{Multiplication:} \)
\[
B=28.41075\,\text{MeV}
\]
\( \textbf{Rounded value:} \)
\[
B\approx28.4\,\text{MeV}
\]
\( \textbf{Final answer:} \) The binding energy is about \(28.4\,\text{MeV}\). Multiplying by \(931.5\), rather than dividing, is required because \(931.5\,\text{MeV}\) corresponds to \(1\,u\) of mass defect.
113. A bound nucleus has binding energy \(B\). If it is completely separated into its individual nucleons, the minimum energy supplied should be:
ⓐ. Less than \(B\), because the nucleons already exist
ⓑ. Equal to \(Bc^2\)
ⓒ. Equal to \(B\) in the ideal minimum case
ⓓ. Zero, because the nucleus is electrically neutral
Correct Answer: Equal to \(B\) in the ideal minimum case
Explanation: Binding energy is defined as the energy needed to separate a nucleus completely into free protons and neutrons. Therefore, in the ideal minimum-energy picture, the supplied energy must equal \(B\). This reverses the formation process, where the same energy is released when free nucleons form the bound nucleus. The factor \(c^2\) is already included when \(B\) is calculated from \(\Delta m c^2\). Electrical neutrality of an atom does not remove the strong nuclear binding inside the nucleus. The value \(B\) refers to the nucleus itself, not only to the net charge of the atom.
114. The table gives four descriptions of mass defect and binding energy. Identify the row with the physically valid statement.
| Row | Description |
|---|
| P | Mass defect is the extra mass gained by a nucleus after binding. |
| Q | Binding energy is the energy equivalent of mass defect. |
| R | Binding energy is caused by electron shells only. |
| S | Mass defect is found by subtracting proton number from mass number. |
ⓐ. Row Q
ⓑ. Row P
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row Q
Explanation: Binding energy is related to mass defect by \(B=\Delta mc^2\). This makes row Q physically valid. Row P is not valid because a bound nucleus has less mass than the separated nucleons, not extra mass. Row R is not valid because nuclear binding is a property of nucleons inside the nucleus, not electron shells. Row S confuses mass defect with neutron-number calculation, since \(A-Z\) gives \(N\), not \(\Delta m\). The correct idea is that a small mass difference corresponds to a large nuclear energy.
115. A binding energy of \(56\,\text{MeV}\) has what mass equivalent in \(u\), using \(1\,u\,c^2=931.5\,\text{MeV}\)?
ⓐ. \(0.120\,u\)
ⓑ. \(0.060\,u\)
ⓒ. \(16.6\,u\)
ⓓ. \(56.0\,u\)
Correct Answer: \(0.060\,u\)
Explanation: \( \textbf{Given:} \) \(B=56\,\text{MeV}\).
\( \textbf{Conversion factor:} \) \(1\,u\,c^2=931.5\,\text{MeV}\).
\( \textbf{Required quantity:} \) Mass equivalent \(\Delta m\) in \(u\).
\( \textbf{Relation:} \)
\[
B=\Delta mc^2
\]
\( \textbf{Conversion setup:} \)
\[
\Delta m=\frac{B}{931.5}\,u
\]
\( \textbf{Substitution:} \)
\[
\Delta m=\frac{56}{931.5}\,u
\]
\( \textbf{Calculation:} \)
\[
\Delta m\approx0.0601\,u
\]
\( \textbf{Final answer:} \) The mass equivalent is approximately \(0.060\,u\). Dividing by \(931.5\) is used here because the direction of conversion is from energy in \(\text{MeV}\) back to mass in \(u\).
116. A claim says, “A larger total binding energy always means a nucleus is more stable than every nucleus with smaller total binding energy.” The most careful response is:
ⓐ. The claim is fully valid because total binding energy alone decides stability across all nuclei
ⓑ. The claim is valid only for atoms with no electrons
ⓒ. The claim is invalid because binding energy is not related to mass defect
ⓓ. Use binding energy per nucleon for stability
Correct Answer: Use binding energy per nucleon for stability
Explanation: Total binding energy usually increases as the number of nucleons increases, because a larger nucleus has more nucleons bound together. However, comparing nuclei of very different sizes using only total binding energy can be misleading. A large nucleus may have a greater total binding energy simply because it contains many nucleons. The average binding per nucleon, \(B/A\), is more useful for comparing stability across different nuclei. Binding energy is still directly related to mass defect by \(B=\Delta mc^2\). The careful stability comparison asks how strongly each nucleon is bound on average.
117. In forming a stable nucleus from separated nucleons, the final nuclear mass becomes smaller. The missing mass appears mainly as:
ⓐ. Extra atomic number
ⓑ. Additional neutrons
ⓒ. A larger value of \(R_0\)
ⓓ. Binding energy
Correct Answer: Binding energy
Explanation: When separated nucleons form a bound nucleus, the system moves to a lower-energy state. The mass of the final bound nucleus is smaller than the total mass of the separated nucleons. The difference in mass is the mass defect. By \(E=\Delta mc^2\), this missing mass corresponds to energy released during formation. The process does not create extra atomic number or additional neutrons. The radius constant \(R_0\) is part of the nuclear size relation and is not the form taken by the mass difference.
118. A nucleus with \(A=8\) has total binding energy \(56\,\text{MeV}\). Its binding energy per nucleon is:
ⓐ. \(8\,\text{MeV}\)
ⓑ. \(48\,\text{MeV}\)
ⓒ. \(7\,\text{MeV}\)
ⓓ. \(64\,\text{MeV}\)
Correct Answer: \(7\,\text{MeV}\)
Explanation: \( \textbf{Given:} \) Total binding energy \(B=56\,\text{MeV}\), and mass number \(A=8\).
\( \textbf{Required quantity:} \) Binding energy per nucleon.
\( \textbf{Relation:} \)
\[
\text{Binding energy per nucleon}=\frac{B}{A}
\]
\( \textbf{Substitution:} \)
\[
\frac{B}{A}=\frac{56\,\text{MeV}}{8}
\]
\( \textbf{Calculation:} \)
\[
\frac{B}{A}=7\,\text{MeV}
\]
\( \textbf{Final answer:} \) The binding energy per nucleon is \(7\,\text{MeV}\). The division by \(A\) is needed because the question asks for the average binding associated with one nucleon.
119. A graph-description question uses this information:
For a set of nuclei, the vertical axis shows total binding energy \(B\), and the horizontal axis shows mass number \(A\). The graph generally rises as \(A\) increases.
Why does this graph alone not directly rank nuclear stability across all nuclei?
ⓐ. Total binding energy hides the nucleon count
ⓑ. Because binding energy has no relation with nuclear forces
ⓒ. Because mass number is not connected with nucleons
ⓓ. Because every nucleus must have the same total binding energy
Correct Answer: Total binding energy hides the nucleon count
Explanation: Total binding energy \(B\) measures the total energy needed to separate the whole nucleus into nucleons. Larger nuclei often have larger \(B\) simply because they contain more nucleons. For stability comparison across different mass numbers, the average value \(B/A\) is more informative. A graph of total \(B\) against \(A\) does not immediately show how strongly each nucleon is bound on average. Binding energy is connected with nuclear forces, but total binding energy alone can hide size effects. Stability comparison usually needs the binding energy per nucleon curve.
120. Consider the statements below.
I. Binding energy is released when separated nucleons form a bound nucleus.
II. The same amount of energy must be supplied to separate the bound nucleus into free nucleons.
III. Binding energy is unrelated to mass defect.
Which statements are valid?
ⓐ. II and III only
ⓑ. I and II only
ⓒ. I only
ⓓ. I, II, and III
Correct Answer: I and II only
Explanation: Statement I is valid because formation of a stable bound nucleus releases energy. Statement II is also valid because separating the nucleus into free nucleons reverses the formation process and requires the binding energy. Statement III is not valid because binding energy is directly related to mass defect by \(B=\Delta mc^2\). The mass defect is the mass equivalent of the binding energy. The formation and separation viewpoints describe the same energy from opposite directions. The sign of energy transfer changes with direction, but the magnitude of the binding energy is the same.
