Physics HOTs for Class 12 with Answers Chapter 13 Nuclei

1. Question:

Calculate the binding energy of a nucleus of deuterium (Hydrogen isotope $^2_1H$ ) if the mass of deuterium is $2.014 \, \text{u}$ , and the masses of the proton and neutron are $1.007 \, \text{u}$ and $1.009 \, \text{u}$ respectively.

Solution: The binding energy of a nucleus is calculated using the mass defect:

\text{Mass defect} = \text{Total mass of nucleons} - \text{Actual mass of nucleus}

For deuterium, which consists of one proton and one neutron:

\text{Mass defect} = (1.007 + 1.009) - 2.014 = 0.002 \, \text{u}

Now, converting the mass defect into energy using Einstein's equation $E = \Delta m c^2$ and converting atomic mass unit (u) to kg ( $1 \, \text{u} = 1.66 \times 10^{-27} \, \text{kg}$ ):

\Delta m = 0.002 \, \text{u} = 0.002 \times 1.66 \times 10^{-27} \, \text{kg} = 3.32 \times 10^{-30} \, \text{kg}

E = (3.32 \times 10^{-30}) \times (3 \times 10^8)^2 = 3.00 \times 10^{-13} \, \text{J}

The binding energy of deuterium is approximately $3.00 \times 10^{-13} \, \text{J}$ .

2. Question:

What is the significance of the mass-energy equivalence in nuclear reactions? Discuss with an example of nuclear fission.

Solution: The mass-energy equivalence principle, expressed by Einstein's equation $E = mc^2$ , states that mass can be converted into energy and vice versa. This is highly relevant in nuclear reactions, such as fission, where a small amount of mass is lost and converted into a large amount of energy.

In nuclear fission, a heavy nucleus (like Uranium-235) splits into two lighter nuclei, releasing a large amount of energy. The mass of the resulting fragments is slightly less than the original mass, and this difference is released as energy:

\text{Energy released} = \Delta m \times c^2

For example, in the fission of Uranium-235, when it absorbs a neutron and splits, the energy released is on the order of 200 MeV per fission event, due to the conversion of mass defect into energy.

3. Question:

Derive the expression for the radius of a nucleus using the liquid drop model.

Solution: According to the liquid drop model, the radius $R$ of a nucleus is given by:

R = R_0 A^{1/3}

where $A$ is the mass number and $R_0$ is a constant.

This relation is derived considering the volume of the nucleus as proportional to the mass number $A$ , assuming the nucleus is spherical and behaves like a drop of liquid. The liquid drop model assumes that the nucleons inside the nucleus are bound by a potential that is similar to the surface tension in a liquid drop. Thus, the radius depends on $A^{1/3}$ , indicating that as the mass number increases, the size of the nucleus increases, but not linearly.

4. Question:

Explain why the binding energy per nucleon increases with the mass number up to iron and then decreases for larger nuclei.

Solution: The binding energy per nucleon increases as the mass number increases for lighter elements because the attractive forces between nucleons become stronger as more nucleons are added, resulting in a more stable nucleus. This is due to the short-range nature of the nuclear force, which is attractive but only effective over small distances.

However, for heavier elements (especially those heavier than iron), the binding energy per nucleon begins to decrease. This is because the repulsive electrostatic forces between protons start to dominate at larger sizes, weakening the binding. As the nucleus gets larger, the nuclear force becomes less effective in holding the nucleus together, and thus the binding energy per nucleon decreases.

5. Question:

A neutron of energy 1 MeV is incident on a Uranium-235 nucleus. Calculate the probability of its fission reaction given that the fission cross-section for Uranium-235 is $2 \times 10^{-24} \, \text{m}^2$ .

Solution: The probability of a neutron causing a fission reaction is proportional to the neutron flux $\phi$ and the fission cross-section $\sigma$ . The rate of fission events is given by:

\text{Rate of fission} = \sigma \times \phi

where $\sigma = 2 \times 10^{-24} \, \text{m}^2$ is the fission cross-section.

If the neutron flux is not given, we assume the probability of a fission event occurring depends on the interaction between the neutron and the uranium nucleus. Since the probability is proportional to the cross-section, for a 1 MeV neutron, the fission reaction probability will be high due to the favorable energy range for Uranium-235 fission.

6. Question:

Calculate the energy released in the fission of Uranium-235 when a single neutron splits the nucleus, assuming the mass defect of the reaction is $0.2 \, \text{u}$ .

Solution: The energy released in a nuclear reaction is given by the mass defect times $c^2$ :

E = \Delta m \times c^2

where $\Delta m = 0.2 \, \text{u} = 0.2 \times 1.66 \times 10^{-27} \, \text{kg} = 3.32 \times 10^{-28} \, \text{kg}$ .

The energy released is:

E = 3.32 \times 10^{-28} \times (3 \times 10^8)^2 = 3.00 \times 10^{-11} \, \text{J}

Converting this energy to MeV:

E = \frac{3.00 \times 10^{-11}}{1.6 \times 10^{-13}} \, \text{MeV} = 187.5 \, \text{MeV}

Thus, the energy released in the fission of Uranium-235 is approximately $187.5 \, \text{MeV}$ .

7. Question:

What is the significance of the concept of nuclear saturation in understanding the nuclear force?

Solution: The concept of nuclear saturation refers to the fact that each nucleon in a nucleus is only strongly interacting with its immediate neighbors and not with all other nucleons in the nucleus. This saturation occurs because the nuclear force is short-range, meaning it only affects nucleons that are close to each other. For example, the force between two nucleons becomes negligible if they are too far apart.

Nuclear saturation explains why the binding energy per nucleon remains relatively constant for medium-sized nuclei (around iron) and why very large nuclei, with more nucleons, require additional forces (like the Coulomb repulsion between protons) to maintain stability.

8. Question:

Calculate the half-life of a substance with a decay constant $\lambda = 1 \times 10^{-3} \, \text{s}^{-1}$ .

Solution: The half-life $T_{1/2}$ of a radioactive substance is related to its decay constant $\lambda$ by the equation:

T_{1/2} = \frac{\ln 2}{\lambda}

Substituting $\lambda = 1 \times 10^{-3} \, \text{s}^{-1}$ :

T_{1/2} = \frac{\ln 2}{1 \times 10^{-3}} = \frac{0.693}{1 \times 10^{-3}} = 693 \, \text{seconds}

Thus, the half-life of the substance is 693 seconds.

9. Question:

Explain the concept of isotopes and how it leads to the existence of different types of nuclear reactions.

Solution: Isotopes are atoms of the same element that have the same number of protons but different numbers of neutrons, and therefore different atomic masses. For example, Uranium-235 and Uranium-238 are isotopes of Uranium with different neutron counts.

The existence of isotopes is significant in nuclear reactions because different isotopes of an element can have vastly different nuclear properties. For example, Uranium-235 is fissile and can undergo fission, while Uranium-238 is not readily fissile but can be converted into Plutonium-239, which is fissile.

This concept leads to different types of nuclear reactions, such as fission, fusion, and neutron capture, depending on the isotope involved.

10. Question:

Explain the role of nuclear reactions in stars, including the process of fusion.

Solution: Nuclear fusion is the process that powers stars. In the cores of stars, under immense pressure and temperature, lighter nuclei, such as hydrogen, fuse together to form heavier nuclei like helium. During this process, a small amount of mass is lost, which is converted into a large amount of energy, according to Einstein’s equation $E = mc^2$ .

In stars like the Sun, hydrogen nuclei (protons) undergo fusion to form helium in a series of reactions known as the proton-proton chain. The energy released during fusion powers the star and provides the light and heat we observe. As the star ages, the fusion process can involve heavier elements like carbon and oxygen.

11. Question:

A nucleus of $^{238}_92U$ undergoes alpha decay. If the energy released in this decay is 4.2 MeV, calculate the kinetic energy of the alpha particle.

Solution: The energy released in a nuclear decay is shared between the decay products (alpha particle and the daughter nucleus). Since the daughter nucleus has much greater mass than the alpha particle, its kinetic energy is almost negligible. Therefore, the energy released in the decay is predominantly carried away by the alpha particle.

So, the kinetic energy of the alpha particle is approximately equal to the energy released:

K.E_{\alpha} = 4.2 \, \text{MeV}

Thus, the kinetic energy of the alpha particle is $4.2 \, \text{MeV}$ .

12. Question:

Derive the expression for the energy of a photon emitted in a transition between two energy levels of a hydrogen atom.

Solution: The energy levels of a hydrogen atom are given by:

E_n = - \frac{13.6}{n^2} \, \text{eV}

where $n$ is the principal quantum number. The energy of a photon emitted when the electron transitions from one energy level to another is the difference between the energies of the two levels:

\Delta E = E_{\text{final}} - E_{\text{initial}} = - \frac{13.6}{n_f^2} + \frac{13.6}{n_i^2}

The energy of the emitted photon is the absolute value of this difference:

E_{\text{photon}} = \frac{13.6}{n_i^2} - \frac{13.6}{n_f^2}

This is the energy of the photon emitted during the transition from level $n_i$ to $n_f$ .

13. Question:

A nucleus of $^{238}_92U$ captures a neutron and undergoes fission. If the fission releases 200 MeV, calculate the total energy released when 1 kg of $^{238}_92U$ undergoes fission.

Solution: The energy released per fission event is 200 MeV. First, we need to calculate the number of $^{238}_92U$ nuclei in 1 kg of Uranium. The mass of one $^{238}_92U$ nucleus is approximately $238 \, \text{u} = 238 \times 1.66 \times 10^{-27} \, \text{kg} = 3.95 \times 10^{-25} \, \text{kg}$ .

The number of nuclei in 1 kg of Uranium is:

N = \frac{1 \, \text{kg}}{3.95 \times 10^{-25} \, \text{kg/nucleus}} = 2.53 \times 10^{24} \, \text{nuclei}

The total energy released is:

\text{Total energy} = N \times 200 \, \text{MeV}

Converting MeV to joules (1 MeV = $1.6 \times 10^{-13} \, \text{J}$ ):

\text{Total energy} = 2.53 \times 10^{24} \times 200 \times 1.6 \times 10^{-13} \, \text{J} = 8.1 \times 10^{13} \, \text{J}

Thus, the total energy released when 1 kg of $^{238}_92U$ undergoes fission is $8.1 \times 10^{13} \, \text{J}$ .

14. Question:

Explain the concept of the nuclear force and its role in the stability of the nucleus.

Solution: The nuclear force is a short-range force that acts between nucleons (protons and neutrons) inside an atomic nucleus. It is much stronger than the electrostatic force at very short distances but has a limited range of approximately 10 fm (femtometers). The nuclear force is attractive at distances up to 3 fm and repulsive at distances less than 0.7 fm.

The role of the nuclear force in the stability of the nucleus is crucial. It binds the protons and neutrons together, overcoming the electrostatic repulsion between positively charged protons. However, for very large nuclei, the balance between the attractive nuclear force and the repulsive electrostatic force becomes delicate. If the nucleus becomes too large, the electrostatic force becomes stronger, leading to instability.

15. Question:

What is the significance of the neutron-to-proton ratio in determining the stability of a nucleus?

Solution: The neutron-to-proton ratio ( $N/Z$ ) is crucial in determining the stability of a nucleus. In stable nuclei, the number of neutrons is approximately equal to the number of protons. For lighter elements (low $Z$ ), the ratio $N/Z$ is close to 1. As elements become heavier (higher $Z$ ), the neutron-to-proton ratio increases slightly, as neutrons are required to counteract the increasing electrostatic repulsion between protons.

If the neutron-to-proton ratio deviates too far from the stable range, the nucleus becomes unstable and undergoes radioactive decay (alpha, beta, or gamma decay) to achieve a more stable configuration. For example, nuclei with too many neutrons undergo beta decay, where a neutron transforms into a proton, while those with too few neutrons undergo beta plus decay.

16. Question:

What is the role of the strong nuclear force in the fission process of heavy nuclei like Uranium-235?

Solution: The strong nuclear force binds nucleons together in the nucleus. During the fission process of heavy nuclei like Uranium-235, a neutron is absorbed by the nucleus, which leads to a distortion in the nuclear structure. This distortion makes the nucleus unstable, and it eventually splits into two smaller nuclei. The fission process is facilitated by the fact that the nuclear force is strong enough to hold the nucleus together at short distances but becomes weaker as the nucleus increases in size. When the nucleus splits, a large amount of energy is released due to the conversion of mass into energy, which is governed by the mass-energy equivalence principle.

17. Question:

Discuss the concept of nuclear reactions in stars and the process of energy production through fusion.

Solution: In stars, energy is primarily produced through nuclear fusion, where lighter elements fuse to form heavier elements. The Sun, for example, primarily undergoes the proton-proton chain reaction, where hydrogen nuclei (protons) combine to form helium, releasing energy in the form of photons.

The energy released in fusion comes from the conversion of mass into energy, as described by Einstein’s equation $E = mc^2$ . The mass of the resulting nucleus is slightly less than the sum of the masses of the individual nuclei, and this mass difference is released as energy. In massive stars, the fusion process can involve heavier elements like carbon and oxygen, ultimately leading to the formation of even heavier elements during supernova explosions.

18. Question:

Explain the concept of radioactivity and discuss the different types of radiation emitted by unstable nuclei.

Solution: Radioactivity is the process by which unstable nuclei lose energy by emitting radiation in the form of particles or electromagnetic waves. There are three main types of radiation:

Alpha decay: Emission of an alpha particle (helium nucleus, $^4_2He$ ) from a nucleus. This reduces the atomic number by 2 and the mass number by 4.
Beta decay: Emission of a beta particle (electron, $\beta^-$ ) and an antineutrino. A neutron decays into a proton, increasing the atomic number by 1 but leaving the mass number unchanged.
Gamma decay: Emission of high-energy electromagnetic radiation (gamma rays) from the nucleus, usually after alpha or beta decay, without changing the atomic or mass number.

These processes are random, but the time it takes for half of a sample to decay is predictable and is known as the half-life.

19. Question:

Calculate the energy required to break up a nucleus of deuterium (mass $2.014 \, \text{u}$ ) into a proton and a neutron, given the binding energy of deuterium is 2.22 MeV.

Solution: The energy required to break a nucleus is equal to the binding energy of the nucleus. For deuterium, the binding energy is 2.22 MeV. This is the energy required to separate the proton and neutron from the nucleus, so the energy required is:

E_{\text{break}} = 2.22 \, \text{MeV}

Hence, the energy required to break up the deuterium nucleus is 2.22 MeV.

20. Question:

Discuss the concept of nuclear fusion as it occurs in the Sun, and explain the proton-proton chain reaction.

Solution: The Sun generates energy through nuclear fusion. In the Sun's core, the temperature and pressure are so high that hydrogen nuclei (protons) collide and fuse, forming helium and releasing energy. The main reaction that occurs is the proton-proton chain.

In the proton-proton chain, four protons (hydrogen nuclei) undergo a series of steps:

Two protons fuse to form deuterium (one proton and one neutron), releasing a positron and a neutrino.
The deuterium nucleus fuses with another proton to form helium-3.
Two helium-3 nuclei fuse to form helium-4 and release two protons.

The net result is that four protons are converted into one helium-4 nucleus, with the release of energy in the form of gamma rays, neutrinos, and positrons.

21. Question:

A sample of Uranium-235 undergoes fission and releases 200 MeV of energy per fission event. How many fission events are required to release 1 GJ of energy?

Solution: Given:

Energy released per fission event = 200 MeV = $200 \times 1.6 \times 10^{-13} \, \text{J} = 3.2 \times 10^{-11} \, \text{J}$
Total energy to be released = 1 GJ = $1 \times 10^9 \, \text{J}$

To find the number of fission events, we use:

\text{Number of fission events} = \frac{\text{Total energy}}{\text{Energy per fission event}} = \frac{1 \times 10^9}{3.2 \times 10^{-11}} = 3.125 \times 10^{19}

Hence, $3.125 \times 10^{19}$ fission events are required to release 1 GJ of energy.

22. Question:

Derive the expression for the binding energy per nucleon for a nucleus.

Solution: The binding energy of a nucleus is the energy required to disassemble the nucleus into its individual nucleons. It is given by:

E_{\text{bind}} = \Delta m \cdot c^2

where $\Delta m$ is the mass defect, which is the difference between the sum of the masses of the individual nucleons and the mass of the nucleus.

The binding energy per nucleon is:

E_{\text{bind per nucleon}} = \frac{E_{\text{bind}}}{A}

where $A$ is the mass number of the nucleus.

Thus, the binding energy per nucleon is directly proportional to the mass defect and is a measure of the stability of the nucleus.

23. Question:

Explain how the concept of mass defect is related to the nuclear binding energy.

Solution: The mass defect is the difference between the mass of a nucleus and the sum of the masses of its constituent nucleons (protons and neutrons). This mass difference arises because some of the mass is converted into energy to hold the nucleus together.

According to Einstein’s mass-energy equivalence $E = \Delta m \cdot c^2$ , the mass defect is directly related to the binding energy of the nucleus. The binding energy is the energy required to break the nucleus into individual protons and neutrons. This energy is what keeps the nucleus stable and is associated with the nuclear force that binds the nucleons together.

24. Question:

If the binding energy of a nucleus is 8 MeV and the mass of the nucleus is 1.67 × 10^-27 kg, calculate the binding energy per nucleon.

Solution: Given:

Binding energy $E_{\text{bind}} = 8 \, \text{MeV}$
Mass of the nucleus $m = 1.67 \times 10^{-27} \, \text{kg}$

First, convert the binding energy into joules:

8 \, \text{MeV} = 8 \times 1.6 \times 10^{-13} \, \text{J} = 1.28 \times 10^{-12} \, \text{J}

The number of nucleons $A$ is given by:

A = \frac{\text{mass of nucleus}}{\text{mass of one nucleon}} = \frac{1.67 \times 10^{-27}}{1.67 \times 10^{-27}} = 1

Therefore, the binding energy per nucleon is $8 \, \text{MeV}$ .

25. Question:

Why are large nuclei unstable? Discuss the factors that contribute to the instability of a nucleus.

Solution: Large nuclei become unstable because the electrostatic repulsion between protons becomes significant as the number of protons increases. This repulsion tends to push the protons apart, while the nuclear force, which binds protons and neutrons together, is short-range and cannot overcome the electrostatic force over large distances.

As the nucleus grows in size, the neutron-to-proton ratio also needs to increase to maintain stability. If the ratio becomes too large or too small, the nucleus becomes unstable and may undergo radioactive decay to achieve a more stable configuration.

26. Question:

In the context of nuclear fission, explain the role of a neutron moderator in a nuclear reactor.

Solution: A neutron moderator is a substance used in nuclear reactors to slow down fast neutrons produced during fission reactions. Fast neutrons have high kinetic energy and are less likely to induce further fission reactions in the reactor fuel. A moderator slows these neutrons down, making them thermal (low-energy) neutrons, which are more likely to cause fission in the fuel (typically uranium-235 or plutonium-239).

Common moderators include water (light or heavy), graphite, and deuterium. By controlling the neutron speed, the moderator helps maintain a controlled chain reaction in the reactor.

27. Question:

Calculate the energy released when 1 mole of Uranium-235 undergoes fission, given that each fission event releases 200 MeV.

Solution: Given:

Energy released per fission event = 200 MeV = $200 \times 1.6 \times 10^{-13} \, \text{J} = 3.2 \times 10^{-11} \, \text{J}$
Number of atoms in 1 mole = $6.022 \times 10^{23}$

The total energy released for 1 mole of Uranium-235 is:

\text{Total energy} = 6.022 \times 10^{23} \times 3.2 \times 10^{-11} \, \text{J} = 1.93 \times 10^{13} \, \text{J}

Hence, the energy released when 1 mole of Uranium-235 undergoes fission is $1.93 \times 10^{13} \, \text{J}$ .

28. Question:

Explain the concept of nuclear magnetic resonance (NMR) and its applications in medicine.

Solution: Nuclear Magnetic Resonance (NMR) is a technique that exploits the magnetic properties of atomic nuclei. It involves placing a sample in a magnetic field and irradiating it with radiofrequency waves. The nuclei absorb the radiofrequency energy and transition between different energy states. The frequency at which this absorption occurs depends on the type of nucleus and its environment.

In medicine, NMR is used in Magnetic Resonance Imaging (MRI). MRI uses the principles of NMR to produce detailed images of the inside of the body. It is a non-invasive imaging technique that helps diagnose a wide range of conditions, such as tumors, brain disorders, and spinal injuries, without the need for ionizing radiation.

29. Question:

Explain the significance of the half-life of a radioactive substance and derive its relation to the decay constant.

Solution: The half-life ( $t_{1/2}$ ) of a radioactive substance is the time required for half of the nuclei in a sample to decay. It is related to the decay constant ( $\lambda$ ), which is the probability per unit time that a nucleus will decay.

The decay law is given by:

N(t) = N_0 e^{-\lambda t}

where $N(t)$ is the number of nuclei remaining after time $t$ , and $N_0$ is the initial number of nuclei.

At half-life $t_{1/2}$ , half of the nuclei decay:

\frac{N_0}{2} = N_0 e^{-\lambda t_{1/2}}

Dividing both sides by $N_0$ , we get:

\frac{1}{2} = e^{-\lambda t_{1/2}}

Taking the natural logarithm of both sides:

\ln \frac{1}{2} = -\lambda t_{1/2}

t_{1/2} = \frac{\ln 2}{\lambda}

Hence, the half-life is inversely proportional to the decay constant.

30. Question:

In the context of nuclear energy, explain the process of nuclear fusion and discuss its advantages over nuclear fission.

Solution: Nuclear fusion is the process in which two light nuclei, typically isotopes of hydrogen (such as deuterium and tritium), combine to form a heavier nucleus, releasing a large amount of energy. The Sun and other stars generate energy through fusion, primarily through the fusion of hydrogen nuclei into helium.

The main advantages of nuclear fusion over nuclear fission include:

Abundant fuel: The fuel for fusion (hydrogen isotopes) is abundant and readily available from water and lithium.
No long-lived radioactive waste: Fusion produces minimal nuclear waste compared to fission, and the waste that is produced is generally less harmful and has a shorter half-life.
No greenhouse gases: Fusion does not produce carbon dioxide or other greenhouse gases during the process.

However, achieving controlled fusion on Earth requires extremely high temperatures and pressures, which has made it technologically challenging. Researchers are working on developing reactors that can sustain the necessary conditions for fusion, such as the ITER project.

Nuclei

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Class 12^th Physics Chapter HOTs