Question bank
Practice A-Level CAIE Physics 23 1 Mass Defect And Nuclear Binding Energy questions by syllabus topic with past-paper context, marks, difficulty and question previews on Eduninja.
Question 2
Question 2(c)
A deuterium nucleus \({ }_{1}^{2} \mathrm{H}\) and a proton collide. A nuclear reaction occurs, represented by the equation
Question 2(c)(i)
State and explain whether the reaction represents nuclear fission or nuclear fusion.
reaction represents either build-up of nucleus from light nuclei or build-up of heavy nucleus from nuclei so fusion reaction A1
Question 4
Question 4(b)
Two deuterium \(\left({ }_{1}^{2} \mathrm{H}\right)\) nuclei each have initial kinetic energy \(E_{\mathrm{K}}\) and are initially separated by a large distance. The nuclei may be considered to be spheres of diameter \(3.8 \times 10^{-15} \mathrm{~m}\) with their masses and charges concentrated at their centres. The nuclei move from their initial positions to their final position of just touching, as illustrated in Fig. 4.1.
Question 4(b)(i)
For the two nuclei approaching each other, calculate the total change in
\(1 \quad \Delta E_{\text {gpe }}=G M m / r\) \(=\left(6.67 \times 10^{-11} \times\left\{2 \times 1.66 \times 10^{-27}\right\}^{2}\right) /\left(3.8 \times 10^{-15}\right)\) \(=1.93 \times 10^{-49} \mathrm{~J}\) \(2 \Delta E_{\text {epe }}=Q q / 4 \pi \varepsilon_{0} r\) \(=\left(1.6 \times 10^{-19}\right)^{2} /\left(4 \pi \times 8.85 \times 10^{-12} \times 3.8 \times 10^{-15}\right)\) C1 \(=6.06 \times 10^{-14} \mathrm{~J}\)
Question 4(b)(ii)
Use your answers in (i) to show that the initial kinetic energy \(E_{\mathrm{K}}\) of each nucleus is 0.19 MeV .
idea that \(2 E_{\mathrm{K}}=\Delta E_{\text {epe }}-\Delta E_{\text {gpe }}\) \(E_{\mathrm{K}}=3.03 \times 10^{-14} \mathrm{~J}\) \(=\left(3.03 \times 10^{-14}\right) / 1.6 \times 10^{-13}\) =0.19 MeV
Question 4(b)(iii)
The two nuclei may rebound from each other. Suggest one other effect that could happen to the two nuclei if the initial kinetic energy of each nucleus is greater than that calculated in (ii).
fusion may occur / may break into sub-nuclear particles
Question 7
A uranium-235 nucleus absorbs a neutron and then splits into two nuclei. A possible nuclear reaction is given by
Question 7(c)
Suggest a possible form of energy released in this reaction.
kinetic energy (of products) or gamma/ \(\gamma\) (radiation or photon) ..... B1 [1]
Question 7(d)
Explain, using the law of mass-energy conservation, how energy is released in this reaction.
(total) mass on left-hand side/reactants is greater than (total) mass on right-hand side/products ..... M1 difference in mass is (converted to) energy ..... A1
Question 7
Question 7(b)
The sum of the masses on the left-hand side of the equation in (a) is not the same as the sum of the masses on the right-hand side. Explain why mass seems not to be conserved.
explanation in terms of mass - energy conservation ..... B1 energy released as gamma or photons or kinetic energy of products or em radiation A1 {\([1]\)} A1 {\([1]\)} A1 {\([1]\)} [2]
Question 39
A slow-moving neutron collides with a nucleus of uranium-235. This results in a nuclear reaction that is represented by the following nuclear equation where x represents one or more particles. What does × represent? one neutron two electrons two neutrons two protons
C
Question 7
A polonium nucleus \({ }_{84}^{210} \mathrm{Po}\) is radioactive and decays with the emission of an \(\alpha\)-particle. The nuclear reaction for this decay is given by
Question 7(a)
Question 7(a)(ii)
Explain why mass seems not to be conserved in the reaction.
mass-energy is conserved B1 mass on rhs is less because energy is released B1 [2]
Question 8
Question 8(a)
State what is meant by the binding energy of a nucleus.
energy required to separate nucleons in a nucleus to infinity
Question 8(b)
Show that the energy equivalence of 1.0 u is 930 MeV .
\(1 \mathrm{u}=1.66 \times 10^{-27} \mathrm{~kg}\)
Question 8(c)
Data for the masses of some particles and nuclei are given in Fig. 8.1. Use data from Fig. 8.1 and information from (b) to determine, in MeV,
Question 8(c)(i)
the binding energy of deuterium, binding energy = MeV
\(\Delta m=2.0141 \mathrm{u}-(1.0073+1.0087) \mathrm{u}\) binding energy \(=1.9 \times 10^{-3} \times 930\)
Question 8(c)(ii)
the binding energy per nucleon of zirconium. Answer all the questions in the spaces provided.
\(\Delta m=(57 \times 1.0087 \mathrm{u})+(40 \times 1.0073 \mathrm{u})-97.0980 \mathrm{u}\) binding energy per nucleon \(=(0.69 \times 930) / 97\)
Question 10
Question 10(a)
Explain what is meant by the binding energy of a nucleus.
energy required to separate the nucleons (in a nucleus) M1 to infinity A1 (allow reverse statement)
Question 10(b)
Data for the masses of some particles are given in Fig. 10.1. The energy equivalent of 1.0 u is 930 MeV .
Question 10(b)(i)
Calculate the binding energy, in MeV , of a tritium \(\left({ }_{1}^{3} \mathrm{H}\right)\) nucleus. binding energy = MeV
\(\Delta m=(2 \times 1.00867)+1.00728-3.01551\) C1 binding energy \(=9.11 \times 10^{-3} \times 930\) (allow 930 to 934 MeV so answer could be in range 8.47 to 8.51 MeV ) (allow 2 s.f.)
Question 10(b)(ii)
The total mass of the separate nucleons that make up a polonium-210 \(\left({ }_{84}^{210} \mathrm{Po}\right)\) nucleus is 211.70394 u. Calculate the binding energy per nucleon of polonium-210. binding energy per nucleon = MeV
\(\Delta m=211.70394-209.93722\) C1 binding energy per nucleon \(=(1.76672 \times 930) / 210\) C1 (allow 930 to 934 MeV so answer could be in range 7.82 to 7.86 MeV ) (allow 2 s.f.)
Question 10(c)
One possible fission reaction is By reference to binding energy, explain, without any calculation, why this fission reaction is energetically possible. Answer all the questions in the spaces provided.
total binding energy of barium and krypton M1 is greater than binding energy of uranium A1
Question 10
Question 10(a)
Explain what is meant by the binding energy of a nucleus.
energy required to separate the nucleons (in a nucleus) M1 to infinity A1 (allow reverse statement)
Question 10(b)
Data for the masses of some particles are given in Fig. 10.1. The energy equivalent of 1.0 u is 930 MeV .
Question 10(b)(i)
Calculate the binding energy, in MeV , of a tritium \(\left({ }_{1}^{3} \mathrm{H}\right)\) nucleus.
\(\Delta m=(2 \times 1.00867)+1.00728-3.01551\) C1 binding energy \(=9.11 \times 10^{-3} \times 930\) (allow 930 to 934 MeV so answer could be in range 8.47 to 8.51 MeV ) (allow 2 s.f.)
Question 10(b)(ii)
The total mass of the separate nucleons that make up a polonium-210 \(\left({ }_{84}^{210} \mathrm{Po}\right)\) nucleus is 211.70394 u. Calculate the binding energy per nucleon of polonium-210. binding energy per nucleon = MeV
\(\Delta m=211.70394-209.93722\) C1 binding energy per nucleon \(=(1.76672 \times 930) / 210\) C1 (allow 930 to 934 MeV so answer could be in range 7.82 to 7.86 MeV ) (allow 2 s.f.)
Question 10(c)
One possible fission reaction is By reference to binding energy, explain, without any calculation, why this fission reaction is energetically possible. Answer all the questions in the spaces provided.
total binding energy of barium and krypton M1 is greater than binding energy of uranium A1
Question 7
Question 7(b)
One possible nuclear reaction involves the bombardment of a stationary nitrogen-14 nucleus by an \(\alpha\)-particle to form oxygen-17 and another particle.
Question 7(b)(ii)
The total mass-energy of the nitrogen-14 nucleus and the \(\alpha\)-particle is less than that of the particles resulting from the reaction. This mass-energy difference is 1.1 MeV . 1. Suggest how it is possible for mass-energy to be conserved in this reaction. 2. Calculate the speed of an \(\alpha\)-particle having kinetic energy of 1.1 MeV . speed = \(\mathrm{m} \mathrm{s}^{-1}\)
1 initially, \(\alpha\)-particle must have some kinetic energy \(\ldots \ldots \ldots \ldots \ldots \ldots .\). B1 (ii) \(21.1 \mathrm{MeV}=1.1 \times 1.6 \times 10^{-13}=1.76 \times 10^{-13} \mathrm{~J} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \quad\) C1 \(1.76 \times 10^{-13}=1 / 2 \times 4 \times 1.66 \times 10^{-27} \times v^{2} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \quad \mathrm{C} 1\) \(v=7.3 \times 10^{6} \mathrm{~m} \mathrm{~s}^{-1} \ldots \ldots \ldots \ldots \ldots . \ldots \ldots . \ldots . . \ldots ~ A1 ~\) [4] use of \(1.67 \times 10^{-27} \mathrm{~kg}\) for mass is a maximum of 3 / 4