Nuclear Physics Archives

21. Which of the following are the properties of an electron? 1. Electron

Which of the following are the properties of an electron?
1. Electron is a constituent of cathode ray
2. Electron is a negatively charged particle
3. The mass of the electron is equal to the mass of the proton
4. Electron is deflected by the electric field but not by magnetic field
Select the correct answer using the code given below:

1 and 2 only

1, 2 and 3

3 and 4

1 and 4

This question was previously asked in

UPSC NDA-2 – 2015

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Statement 1 is correct: Cathode rays are streams of electrons. Statement 2 is correct: Electrons are negatively charged fundamental particles. Statement 3 is incorrect: The mass of an electron is significantly smaller (about 1/1836) than the mass of a proton. Statement 4 is incorrect: Electrons, being charged particles, are deflected by both electric fields and magnetic fields.

This question tests basic properties of electrons, a fundamental particle in physics and chemistry.

Electrons are leptons and are one of the constituents of atoms, orbiting the nucleus. Their charge is approximately -1.602 x 10⁻¹⁹ Coulombs.

22. Of the following, which does not belong to a nuclear reactor ?

Of the following, which does not belong to a nuclear reactor ?

A turbine

A heat exchanger

A mechanism to reduce CO2 emission

A reaction chamber

This question was previously asked in

UPSC NDA-1 – 2024

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A mechanism to reduce CO2 emission does not belong to a nuclear reactor itself.

Nuclear reactors are power plants that generate electricity through nuclear fission, a process that does not produce carbon dioxide. Therefore, a mechanism *within the reactor* specifically designed to reduce CO2 emissions is not a component of a nuclear reactor. Other listed items (turbine, heat exchanger, reaction chamber/core) are integral parts of the nuclear power generation process.

Nuclear power is often promoted as a low-carbon energy source because it avoids the CO2 emissions associated with burning fossil fuels. However, this is a characteristic of the energy source itself (absence of CO2 production during fission) rather than a component within the reactor designed for CO2 reduction.

23. Which one of the following conclusions could not be derived from Rut

Which one of the following conclusions could not be derived from Rutherford’s α-particle scattering experiment ?

Most of the space in the atom is empty.

The radius of the atom is about 10⁵ times the radius of the nucleus.

Electrons move in a circular path of fixed energy called orbits.

Nearly all the mass of the atom resides in the nucleus.

This question was previously asked in

UPSC NDA-1 – 2021

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Rutherford’s alpha-particle scattering experiment (Gold foil experiment) led to the discovery of the atomic nucleus and the proposal of the nuclear model of the atom. The key conclusions were:
A) Most alpha particles passed straight through, indicating that most of the atom is empty space.
D) A small fraction of particles were deflected at large angles or bounced back, indicating the presence of a dense, positively charged nucleus containing most of the mass in a tiny volume.
B) Based on the trajectories, Rutherford estimated the size of the nucleus relative to the atom.
However, Rutherford’s model did not explain the stability of the atom (electrons orbiting a nucleus should lose energy and spiral into the nucleus) nor the observed atomic spectra. The idea that electrons move in specific circular paths of fixed energy (orbits) was a postulate introduced later by Niels Bohr to address these issues, building upon Rutherford’s model. Therefore, this conclusion was not derived from Rutherford’s experiment.

Rutherford’s model established the concept of a nucleus but did not explain electron behaviour in detail. Bohr’s model later quantized electron energy levels and introduced the concept of stable orbits.

The planetary model is often associated with Rutherford, but the idea of quantized orbits is specifically from Bohr. The scattering experiment involved firing alpha particles (He²⁺ ions) at a thin gold foil and observing their deflection patterns.

24. Basic scientific principle behind a nuclear reactor is

Basic scientific principle behind a nuclear reactor is

Nuclear fusion

Controlled nuclear fusion

Uncontrolled nuclear fission

Controlled nuclear fission

This question was previously asked in

UPSC NDA-1 – 2019

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The correct answer is (D) Controlled nuclear fission.

Nuclear reactors generate energy by splitting heavy atomic nuclei, such as Uranium-235, in a process called nuclear fission. This process releases a large amount of energy, neutrons, and other particles.

In a nuclear reactor, the fission chain reaction is carefully controlled using materials like control rods (which absorb neutrons) and moderators (which slow down neutrons) to ensure a sustained but non-explosive release of energy. This controlled reaction differentiates a nuclear reactor from an atomic bomb, which relies on an uncontrolled chain reaction (uncontrolled nuclear fission). Nuclear fusion, the process of combining light nuclei, is the principle behind the sun’s energy and thermonuclear weapons, but it is not yet used for commercial power generation in current nuclear reactors due to the extremely high temperatures and pressures required.

25. Which one of the following minerals is used as a fuel in nuclear power

Which one of the following minerals is used as a fuel in nuclear power stations ?

Bauxite

Quartz

Feldspar

Pitchblende

This question was previously asked in

UPSC NDA-1 – 2019

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The correct answer is D) Pitchblende.

Nuclear power stations use fissile materials, primarily uranium, as fuel for nuclear fission. Pitchblende is a mineral that is a major ore of uranium. Bauxite is the main ore of aluminium. Quartz is a common mineral made of silicon dioxide. Feldspar is a group of rock-forming silicate minerals. Therefore, Pitchblende is the mineral associated with the fuel used in nuclear power stations.

Uranium is extracted and processed from uranium ores like pitchblende (Uraninite) to produce fuel rods for nuclear reactors. The specific isotope used for fission is Uranium-235.

26. Rutherford’s alpha-particle scattering experiment was responsible for

Rutherford’s alpha-particle scattering experiment was responsible for the discovery of

Electron

Proton

Nucleus

Helium

This question was previously asked in

UPSC NDA-1 – 2017

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Rutherford’s alpha-particle scattering experiment, also known as the gold foil experiment, involved firing alpha particles (positively charged helium nuclei) at a thin gold foil. The observation that most particles passed through, some were deflected, and a few were scattered back at large angles led Rutherford to propose that atoms have a very small, dense, positively charged center, which he named the nucleus.

Rutherford’s alpha-scattering experiment established the existence of the atomic nucleus.

While the experiment provided evidence for a positive charge in the nucleus (later identified as protons), it was the discovery of the nucleus itself as the dense core of the atom that was the direct result of analyzing the scattering patterns. Electrons were discovered earlier by J.J. Thomson. Neutrons were discovered later by James Chadwick.

27. Nucleus ²⁴⁰U, which has a binding energy per nucleon as 7·6 MeV, disin

Nucleus ²⁴⁰U, which has a binding energy per nucleon as 7·6 MeV, disintegrates into two nuclei of ¹¹⁹⋅⁵Sn. Take ¹¹⁹⋅⁵Sn elements atomic mass number as 120 and binding energy per nucleon as 8·4 MeV. Which one among the following is the correct value of the energy released in the disintegration process ?

192 MeV

190 MeV

188 MeV

3840 MeV

This question was previously asked in

UPSC Geoscientist – 2024

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Correct Answer: A

– The initial nucleus is ²⁴⁰U with A=240 and binding energy per nucleon = 7.6 MeV. Total binding energy of ²⁴⁰U = 240 * 7.6 MeV.
– The nucleus disintegrates into two nuclei of ¹¹⁹⋅⁵Sn. We are given to use the atomic mass number as 120 for calculation. So, the two product nuclei have A=120 and binding energy per nucleon = 8.4 MeV.
– Total binding energy of the two product nuclei = 2 * (120 * 8.4 MeV).
– The energy released in a nuclear reaction is the difference between the total binding energy of the products and the total binding energy of the reactants. Energy released = (Total binding energy of products) – (Total binding energy of reactants).
– Energy released = (2 * 120 * 8.4 MeV) – (240 * 7.6 MeV)
– Energy released = (240 * 8.4 MeV) – (240 * 7.6 MeV)
– Energy released = 240 * (8.4 – 7.6) MeV
– Energy released = 240 * 0.8 MeV
– Energy released = 192 MeV.

Binding energy represents the energy required to separate the nucleons (protons and neutrons) in a nucleus. A higher binding energy per nucleon indicates a more stable nucleus. In nuclear reactions like fission or fusion, energy is released when the products are more stable (have higher binding energy per nucleon) than the reactants. The mass difference between the reactant and product nuclei (mass defect) is converted into energy according to Einstein’s famous equation E=mc². The energy released can also be calculated from the difference in binding energies, as shown here.

28. What is the use of heavy water?

What is the use of heavy water?

It is used in nuclear reactors as a moderator

It is used in nuclear reactors as fuel

It is used in radiation therapy for cancer

It is used in water softening plants

This question was previously asked in

UPSC Geoscientist – 2023

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Heavy water (D₂O) is deuterium oxide, where hydrogen atoms are replaced by deuterium, an isotope of hydrogen with one proton and one neutron. In nuclear reactors, heavy water is primarily used as a moderator. A moderator slows down the fast neutrons released during nuclear fission to thermal neutron energies, which are required to sustain a nuclear chain reaction, especially in reactors using natural uranium as fuel.

Heavy water acts as a neutron moderator in certain types of nuclear reactors, effectively slowing down neutrons without absorbing them significantly.

Besides being a moderator, heavy water can also serve as a coolant in nuclear reactors. Unlike light water reactors (using H₂O), heavy water reactors can often use unenriched uranium as fuel. While nuclear reactors produce radioisotopes used in cancer therapy, heavy water itself is not used directly in radiation therapy. Water softening typically involves removing hard ions like Ca²⁺ and Mg²⁺, not a function of heavy water.

29. Which one of the following statements for the emission spectrum of hyd

Which one of the following statements for the emission spectrum of hydrogen is true?

The Lyman series lies in the visible region and the Paschen series lies in the infrared region.

The Lyman series lies in the ultraviolet region and the Paschen series lies in the infrared region.

Both the Lyman and the Paschen series lie in the visible region.

The Lyman series lies in the ultraviolet region and the Paschen series lies in the visible region.

This question was previously asked in

UPSC Geoscientist – 2023

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The emission spectrum of hydrogen arises from the de-excitation of electrons from higher energy levels (n_initial) to lower energy levels (n_final). Different series are named based on the final energy level (n_final).
– Lyman series: n_final = 1. Transitions from n_initial = 2, 3, 4,… to n=1. These transitions involve the largest energy drops and thus result in the emission of high-energy photons, which fall in the ultraviolet (UV) region of the electromagnetic spectrum.
– Balmer series: n_final = 2. Transitions from n_initial = 3, 4, 5,… to n=2. These transitions correspond to visible light.
– Paschen series: n_final = 3. Transitions from n_initial = 4, 5, 6,… to n=3. These transitions involve smaller energy drops than Balmer series and fall in the infrared (IR) region of the electromagnetic spectrum.
– Brackett series: n_final = 4. Transitions from n_initial = 5, 6, 7,… to n=4. These are in the far-infrared region.
– Pfund series: n_final = 5. Transitions from n_initial = 6, 7, 8,… to n=5. These are also in the far-infrared region.
The statement that is true is that the Lyman series lies in the ultraviolet region and the Paschen series lies in the infrared region.

Different series in the hydrogen emission spectrum are defined by the principal quantum number of the final energy level of the electron transition. The energy and thus the region of the spectrum depend on the energy difference between the initial and final levels.

The energy of the emitted photon is given by the difference in energy between the initial and final states, E = R_H * (1/n_final² – 1/n_initial²), where R_H is the Rydberg constant. Larger energy differences (smaller n_final) correspond to shorter wavelengths (higher energy photons).

30. Which one of the following statements is true for nuclear fission

Which one of the following statements is true for nuclear fission processes?

Light initial nuclides have lesser number of neutrons than protons.

Heavy initial nuclides have lesser number of neutrons than protons.

All initial nuclides have larger number of neutrons than protons.

All initial nuclides have equal number of protons and neutrons.

This question was previously asked in

UPSC Geoscientist – 2022

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Nuclear fission involves the splitting of heavy atomic nuclei, such as isotopes of Uranium or Plutonium. For stable nuclei, as the atomic number (number of protons, Z) increases, the ratio of neutrons (N) to protons increases beyond 1 to compensate for the increasing electrostatic repulsion between protons. Heavy nuclei that undergo fission are located well above the line of stability on an N-Z chart and are typically neutron-rich. For example, Uranium-235 has 92 protons and 143 neutrons (N/Z ≈ 1.55), and Uranium-238 has 92 protons and 146 neutrons (N/Z ≈ 1.59). Thus, the heavy initial nuclides involved in fission processes have a larger number of neutrons than protons.

– Fission involves heavy nuclei.
– Heavy stable and fissionable nuclei have N > Z.
– The neutron-to-proton ratio increases with increasing atomic number for stable nuclei.

The fission process itself typically releases neutrons. This is because the fission fragments (the lighter nuclei produced) are more stable with a lower N/Z ratio than the initial heavy nucleus, so excess neutrons are emitted during the process, which can sustain a chain reaction.