Modern Physics – NDA Physics PYQs

NDA 2025-II

Q. 1. The wavelength of X-rays is of the order of

(a)  1 Å

(b)  1 cm

(c)  1 mm

(d)  1 m

Answer: (a) 1 Å

Explanation: X-rays have wavelengths in the range ~0.01 nm to 10 nm = 0.1 Å to 100 Å. 1 Å (= 10⁻¹⁰ m = 0.1 nm) is right in the middle of the X-ray range. 1 cm, 1 mm, and 1 m are all in the microwave to radio wave range. Concept Tested: X-ray wavelength: order of 1 Å (= 10⁻¹⁰ m = 0.1 nm)

Q. 2. The dimension of Planck’s constant h is the same as that of

(a)  energy

(b)  angular momentum

(c)  momentum

(d)  power

Answer: (b) angular momentum

Explanation: From E = hf: h = E/f has units of J/Hz = J·s = kg·m²·s⁻¹. Angular momentum L = mvr = m·(m/s)·m = kg·m²·s⁻¹. Both h and angular momentum have the same dimensions [M L² T⁻¹]. Linear momentum has dimensions [M L T⁻¹] (different). Energy is [M L² T⁻²]; power is [M L² T⁻³]. Concept Tested: Planck’s constant dimensions: [M L² T⁻¹] = same as angular momentum (J·s)

NDA 2024-I

Q. 3. Which one of the following proposed that electrons revolve around the nucleus in stable orbits without radiating any energy?

(a)  Rutherford

(b)  Bohr

(c)  J.J. Thomson

(d)  Einstein

Answer: (b) Bohr

Explanation: Niels Bohr (1913) proposed that electrons revolve in specific stable orbits (stationary states) around the nucleus without emitting radiation. Radiation is emitted only when an electron jumps between orbits. This added the quantum orbital model to Rutherford’s nuclear model. Concept Tested: Bohr’s atomic model: electrons in stable orbits; no radiation while orbiting

Q. 4. Which one of the following is NOT a component of a nuclear power plant?

(a)  Reaction chamber

(b)  Heat exchanger

(c)  Control rods

(d)  CO₂ emission reducer

Answer: (d) CO₂ emission reducer

Explanation: A nuclear power plant contains: a reaction chamber (core, where fission occurs), a heat exchanger (transfers heat to steam), control rods (absorb neutrons, regulate the chain reaction), a moderator, and turbines. A CO₂ emission reducer is associated with fossil fuel power stations, not nuclear plants. Concept Tested: Nuclear power plant components: reaction chamber, heat exchanger, control rods; NOT CO₂ reducer

NDA 2023-I

Q. 5. X-rays can be used for

(a)  studying crystalline structure

(b)  radar

(c)  both studying crystalline structure and radar

(d)  neither studying crystalline structure nor radar

Answer: (a) studying crystalline structure

Explanation: X-rays (wavelength ~0.1–10 nm) are used for studying crystal structure by X-ray diffraction (the spacing of crystal planes is comparable to X-ray wavelengths). X-rays are also used in medical imaging and radiation therapy. Radar uses radio waves (centimetre-wavelength microwaves): not X-rays, which penetrate rather than reflect. Concept Tested: X-ray uses: crystallography, medical imaging, radiation therapy; NOT radar (radar uses radio/microwaves)

★ JOVIK Exam Insight X-rays are NOT used for radar. Radar requires waves that reflect off objects and atmospheric layers: radio waves and microwaves. X-rays are used for crystallography (Bragg diffraction) where wavelength ≈ crystal plane spacing.

NDA 2022-II

Q. 6. A radiation of wavelength 1 nm is classified as

(a)  Radio wave

(b)  X-ray

(c)  Infrared

(d)  Ultraviolet

Answer: (b) X-ray

Explanation: 1 nm = 10 Å. X-ray wavelength range: 0.01–10 nm. 1 nm is well within the X-ray range. UV starts at about 10 nm; visible at 380 nm; infrared beyond 780 nm. Concept Tested: Wavelength 1 nm: classified as X-ray

Q. 7. Which one of the following converts light energy to electrical energy?

(a)  LED

(b)  Solar cell

(c)  Laser diode

(d)  Transistor

Answer: (b) Solar cell

Explanation: A solar cell (photovoltaic cell) converts light energy directly into electrical energy through the photovoltaic effect: a form of the photoelectric effect in semiconductors. An LED converts electrical energy to light. A laser diode also converts electrical energy to coherent light. A transistor is a semiconductor device for amplifying or switching electrical signals. Concept Tested: Solar cell: converts light energy to electrical energy (photovoltaic effect)

NDA 2021-II

Q. 8. The discovery of nucleus was made by

(a)  J.J. Thomson

(b)  Ernest Rutherford

(c)  Niels Bohr

(d)  Albert Einstein

Answer: (b) Ernest Rutherford

Explanation: Ernest Rutherford discovered the atomic nucleus in 1911 through the gold foil alpha-particle scattering experiment. J.J. Thomson discovered the electron (1897). Niels Bohr proposed the planetary atomic model with electron orbits. Einstein worked on relativity and the photoelectric effect. Concept Tested: Discovery of atomic nucleus: Ernest Rutherford (1911 gold foil experiment)

Q. 9. The expression E = mc² represents energy released in nuclear reactions. This expression was proposed by

(a)  Rutherford

(b)  Bohr

(c)  Einstein

(d)  Heisenberg

Answer: (c) Einstein

Explanation: E = mc² is Einstein’s mass-energy equivalence from special relativity (1905). It quantifies the energy equivalent of a given mass: the basis for understanding why nuclear reactions (fission and fusion) release such enormous energy. Concept Tested: E = mc²: proposed by Einstein (mass-energy equivalence from special relativity)

NDA 2021-I

Q. 10. Consider the following facts about Rutherford’s alpha-particle scattering experiment:

1.  Most of the atom is empty space

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

3.  Radius of the atom is about 10⁵ times the radius of the nucleus

4.  Electrons move in circular orbits of fixed energy around the nucleus

Which of the above statements are correct conclusions of Rutherford’s experiment?

(a)  1, 2, 3 and 4

(b)  1, 2 and 3 only

(c)  3 and 4 only

(d)  2 and 4 only

Answer: (b) 1, 2 and 3 only

Explanation: Statements 1, 2, and 3 are all correct conclusions from the alpha-particle scattering experiment. Statement 4 (electrons in circular orbits of fixed energy) was proposed by Niels Bohr in his 1913 model: it was NOT a conclusion of Rutherford’s experiment. Concept Tested: Rutherford’s experiment: conclusions: empty space, mass in nucleus, atom 10⁵ × nucleus. Bohr added electron orbits.

Q. 11. Rutherford’s gold foil experiment led to the discovery of the

(a)  electron

(b)  neutron

(c)  nucleus

(d)  proton

Answer: (c) nucleus

Explanation: Rutherford’s alpha-particle scattering experiment led to the discovery of the atomic nucleus: a tiny, dense, positively charged core. J.J. Thomson discovered the electron; James Chadwick discovered the neutron; Rutherford also discovered the proton but through a different experiment (nitrogen bombardment). Concept Tested: Rutherford’s gold foil experiment: discovered the atomic nucleus

NDA 2020-I & II

Q. 12. The shortest wavelength among the following is for

(a)  Infrared rays

(b)  Ultraviolet rays

(c)  X-rays

(d)  Microwaves

Answer: (c) X-rays

Explanation: In order of increasing wavelength: X-ray < UV < visible < infrared < microwave < radio. X-rays have the shortest wavelength among the four options (approximately 0.01–10 nm). Concept Tested: EM spectrum: X-rays have shortest wavelength among IR, UV, X-ray, microwave

NDA 2019-II

Q. 13. In a cathode ray tube, cathode rays travel

(a)  from cathode to anode

(b)  from anode to cathode

(c)  from cathode to anode and back

(d)  randomly in the tube

Answer: (a) from cathode to anode

Explanation: Cathode rays are streams of electrons emitted from the cathode (negative electrode) and accelerated toward the anode (positive electrode). They travel from cathode to anode: in the direction of electron flow (negative to positive). Conventional current is opposite to electron flow. Concept Tested: Cathode rays: travel from cathode (−) to anode (+); direction of electron flow

Q. 14. Cathode rays were deflected by

(a)  electric field only

(b)  magnetic field only

(c)  both electric and magnetic fields

(d)  neither electric nor magnetic fields

Answer: (c) both electric and magnetic fields

Explanation: Cathode rays consist of electrons: charged particles. Charged particles are deflected by both electric fields (Coulomb force) and magnetic fields (Lorentz force). J.J. Thomson used deflection in both types of fields to measure the charge-to-mass ratio (e/m) of the electron. Concept Tested: Cathode rays deflected by: both electric AND magnetic fields (charged particles)

★ JOVIK Exam Insight A common NDA confusion: cathode rays are deflected only by electric fields. Wrong. Thomson used BOTH types of fields in his experiment to determine e/m for the electron.

NDA 2019-I

Q. 15. A nuclear reactor works on the principle of

(a)  nuclear fission

(b)  nuclear fusion

(c)  radioactive decay

(d)  nuclear fission as well as nuclear fusion

Answer: (a) nuclear fission

Explanation: A nuclear reactor operates on controlled nuclear fission: heavy nuclei (U-235 or Pu-239) are split by slow neutrons, releasing heat used to generate electricity. Fusion requires extreme temperatures and pressures not yet achievable in a controlled sustained manner for power generation. Radioactive decay releases far less energy. Concept Tested: Nuclear reactor: controlled nuclear fission of U-235 or Pu-239

Q. 16. The mineral from which uranium is extracted is

(a)  Bauxite

(b)  Quartz

(c)  Pitchblende

(d)  Feldspar

Answer: (c) Pitchblende

Explanation: Pitchblende (uraninite, UO₂) is the primary ore of uranium. Marie Curie discovered polonium and radium from pitchblende residues. Bauxite is aluminium ore; quartz is SiO₂; feldspar is a silicate mineral. Concept Tested: Uranium ore: pitchblende (uraninite); discovered by Marie Curie

Q. 17. For which one of the following was the Nobel prize awarded to Albert Einstein?

(a)  Special theory of relativity

(b)  Quantum theory of light and photoelectric effect

(c)  General theory of relativity

(d)  Kinetic theory of gases

Answer: (b) Quantum theory of light and photoelectric effect

Explanation: Einstein was awarded the Nobel Prize in Physics in 1921 for his discovery of the law of the photoelectric effect: explaining it using Planck’s quantum theory by proposing light quanta (photons). He was famously NOT awarded the Nobel for relativity. Concept Tested: Einstein’s Nobel Prize: photoelectric effect and quantum theory of light (not relativity)

NDA 2018-II

Q. 18. Consider the following statements about visible light, ultraviolet and X-rays:

1.  Wavelength: visible > ultraviolet > X-ray

2.  Photon energy: X-ray > ultraviolet > visible

3.  Energy of ultraviolet photon < energy of visible light photon

Which of the statements given above are correct?

(a)  1 and 3

(b)  2 and 3

(c)  1 and 2

(d)  1, 2 and 3

Answer: (c) 1 and 2

Explanation: Statement 1 ✓: visible light (400–780 nm) has longer wavelength than UV (10–380 nm) which has longer wavelength than X-ray (0.01–10 nm). Statement 2 ✓: photon energy increases as wavelength decreases: so X-ray > UV > visible. Statement 3 ✗: UV photon energy is GREATER than visible photon energy (shorter wavelength = more energy per photon). Statement 3 is false. Concept Tested: EM spectrum comparison: wavelength: visible > UV > X-ray; energy: X-ray > UV > visible

Q. 19. The wavelength of X-rays is of the order of

(a)  100 nm

(b)  1 nm

(c)  500 nm

(d)  5000 nm

Answer: (b) 1 nm

Explanation: X-ray wavelengths are approximately 0.01–10 nm (1 Å = 0.1 nm to 100 Å = 10 nm). 1 nm is in the middle of the X-ray range. 100 nm is UV; 500 nm is visible light (green); 5000 nm is infrared. Concept Tested: X-ray wavelength: order of 1 nm (0.01–10 nm range)

NDA 2017-II

Q. 20. The ionisation energy of the hydrogen atom in the ground state is

(a)  13.6 eV

(b)  13.6 MeV

(c)  13.6 joules

(d)  zero

Answer: (a) 13.6 eV

Explanation: The ionisation energy of hydrogen is 13.6 eV: the energy required to remove the electron from the ground state (n = 1) to infinity. 1 MeV = 10⁶ eV (nuclear scale); 13.6 joules would be ~8.5 × 10¹⁹ eV. The correct scale is electron-volts (eV). Concept Tested: Hydrogen ionisation energy: 13.6 eV (not 13.6 MeV: that is 10⁶ times too large)

★ JOVIK Exam Insight Common error: 13.6 MeV. This is wrong by a factor of 10⁶. The eV scale applies to atomic electron energies. The MeV scale applies to nuclear binding energies. Atomic ionisation energies are always in eV range.

Q. 21. Among the following, which one has the maximum energy per photon?

(a)  Radio wave

(b)  Light wave

(c)  Microwave

(d)  X-ray

Answer: (d) X-ray

Explanation: Photon energy E = hf = hc/λ. Shorter wavelength (higher frequency) = higher photon energy. In order: radio < microwave < visible light < X-ray. X-rays have the shortest wavelength and highest frequency among the options, so they carry the maximum energy per photon. Concept Tested: Maximum photon energy: X-ray (shortest wavelength, highest frequency, highest energy per photon)

Q. 22. The Nobel Prize for explaining the photoelectric effect was awarded to

(a)  Max Planck

(b)  Albert Einstein

(c)  Niels Bohr

(d)  Ernest Rutherford

Answer: (b) Albert Einstein

Explanation: Albert Einstein received the Nobel Prize in Physics in 1921 for explaining the photoelectric effect using Planck’s quantum hypothesis: proposing that light consists of photons, each with energy E = hf. Max Planck received the 1918 Nobel for proposing quantisation of energy. Bohr received it for the atomic model; Rutherford for nuclear chemistry. Concept Tested: Photoelectric effect: explained by Albert Einstein (Nobel 1921); not Planck, Bohr, or Rutherford

Q. 23. The device used to measure radioactivity is

(a)  Polarimeter

(b)  Calorimeter

(c)  Geiger-Müller counter

(d)  Colorimeter

Answer: (c) Geiger-Müller counter

Explanation: A Geiger-Müller (GM) counter detects ionising radiation by counting electrical pulses produced when radiation ionises gas in a sealed tube. A polarimeter measures rotation of polarised light. A calorimeter measures heat energy. A colorimeter measures light absorption. Concept Tested: Geiger-Müller counter: measures radioactivity (detects ionising radiation)

NDA 2017-I

Q. 24. Which one of the following experiments led to the discovery of the nucleus of atom?

(a)  Photoelectric effect experiment

(b)  Rutherford’s alpha-particle scattering experiment

(c)  Compton effect experiment

(d)  Thomson’s cathode ray experiment

Answer: (b) Rutherford’s alpha-particle scattering experiment

Explanation: Rutherford’s 1909–1911 experiment with alpha particles fired at a gold foil revealed that most alpha particles passed straight through, but a small fraction deflected at large angles or bounced back. This led Rutherford to propose that the atom has a tiny, dense, positively charged nucleus. J.J. Thomson’s cathode ray experiment discovered the electron (1897), not the nucleus. Concept Tested: Rutherford’s alpha-particle scattering: discovered the atomic nucleus (not electron, not photoelectric effect)

★ JOVIK Exam Insight Three NDA papers test Rutherford vs Thomson confusion: 2017-I, 2021-I, 2021-II. Always: Rutherford → nucleus. Thomson → electron. These are separate experiments by separate scientists.

Q. 25. If the accelerating potential difference for the electrons in an X-ray tube is doubled, then the cutoff wavelength of X-rays will become

(a)  double

(b)  four times

(c)  half

(d)  one-fourth

Answer: (c) half

Explanation: The cutoff (minimum) wavelength: λ_min = hc/(eV). V ∝ 1/λ_min. If V doubles, λ_min halves. More voltage → more energy → shorter minimum wavelength. Concept Tested: Cutoff wavelength: λ_min = hc/eV; doubling voltage halves λ_min

NDA 2015-II

Q. 26. A radiation of wavelength about 1 nanometre (nm) is called

(a)  infrared radiation

(b)  radio wave

(c)  X-ray

(d)  ultraviolet radiation

Answer: (c) X-ray

Explanation: X-rays have wavelengths in the range ~0.01–10 nm (0.1–100 Å). A wavelength of 1 nm falls squarely in the X-ray range. Infrared is 780 nm–1 mm; radio waves are > 1 mm; UV is 10–380 nm. Concept Tested: Wavelength 1 nm: falls in X-ray range (0.01–10 nm)

★ JOVIK Exam Insight X-ray wavelength of ~1 nm tested in 4 papers: 2015-II, 2018-II, 2022-II, 2025-II. Always: X-ray wavelength ≈ 1 Å to 10 nm.

Q. 27. Which one of the following statements about the electron is NOT correct?

(a)  Electron is a negatively charged particle

(b)  The mass of electron is equal to the mass of proton

(c)  Electrons are the constituents of cathode rays

(d)  Electrons can be deflected by electric field

Answer: (b) The mass of electron is equal to the mass of proton

Explanation: Option (b) is false: the electron mass (9.11 × 10⁻³¹ kg) is approximately 1/1836 of the proton mass (1.67 × 10⁻²⁷ kg). Options (a), (c), and (d) are all correct: electrons are negative, constitute cathode rays, and are deflected by electric fields (and also by magnetic fields). Concept Tested: Electron properties: mass ≠ proton mass (m_e = m_p/1836); rest are correct

Q. 28. Consider the following statements about atomic number:

1.  It is equal to the number of protons in the nucleus.

2.  It is equal to the number of electrons in the shells of the atom.

3.  It is equal to the total number of protons and neutrons in the nucleus.

Which of the statements given above are correct?

(a)  1, 2 and 3

(b)  1 and 2 only

(c)  2 and 3 only

(d)  1 and 3 only

Answer: (b) 1 and 2 only

Explanation: Atomic number Z = number of protons in nucleus (1 ✓). For a neutral atom, Z also equals the number of electrons in the shells (2 ✓). Statement 3 is wrong: the total number of protons + neutrons is the mass number (A), not the atomic number. Concept Tested: Atomic number: equals protons AND electrons (neutral atom); NOT protons + neutrons (that’s mass number)

NDA 2013-I

Q. 29. The energy of the Sun primarily comes from

(a)  nuclear fission

(b)  nuclear fusion

(c)  gravitational energy

(d)  chemical energy

Answer: (b) nuclear fusion

Explanation: The Sun converts hydrogen to helium through nuclear fusion at its core. Four hydrogen nuclei (protons) fuse to form one helium nucleus, releasing energy as the helium mass is slightly less than the combined hydrogen mass: the mass defect is converted to energy via E = mc². Concept Tested: Sun’s energy: nuclear fusion (hydrogen → helium); E = mc²

Q. 30. What does an ionisation chamber measure?

(a)  Intensity of sound waves

(b)  Intensity of radiation

(c)  Strength of magnetic field

(d)  Pressure of gas

Answer: (b) Intensity of radiation

Explanation: An ionisation chamber detects and measures the intensity of ionising radiation (X-rays, gamma rays, alpha and beta particles) by measuring the ionisation current produced when radiation passes through the gas inside the chamber. It is used in medical physics and radiation protection. Concept Tested: Ionisation chamber: measures intensity of ionising radiation

NDA 2012-II

Q. 31. Which one of the following statements about the Sun is correct?

(a)  The Sun derives its energy from nuclear fission of heavier elements

(b)  The Sun derives its energy from nuclear fusion of lighter elements

(c)  The temperature at the surface of the Sun is about 6 × 10⁷ K

(d)  The Sun derives its energy from chemical reactions

Answer: (b) The Sun derives its energy from nuclear fusion of lighter elements

Explanation: The Sun fuses hydrogen (lightest element) into helium at temperatures of ~1.5 × 10⁷ K at the core. This is nuclear fusion of lighter elements. The surface temperature is approximately 5778 K (~6000 K): not 6 × 10⁷ K (that’s the core temperature). Chemical reactions produce far too little energy. Concept Tested: Sun’s energy: nuclear fusion of lighter elements (hydrogen → helium); surface T ≈ 6000 K

NDA 2012-I

Q. 32. In an X-ray tube, the penetrating power of X-rays increases when the

(a)  filament current increases

(b)  filament current decreases

(c)  potential difference between cathode and anode increases

(d)  potential difference between cathode and anode decreases

Answer: (c) potential difference between cathode and anode increases

Explanation: Penetrating power depends on X-ray energy. Higher potential difference (voltage) → electrons accelerated to higher kinetic energy → X-rays of higher energy (shorter wavelength, more penetrating). Filament current controls the intensity (number) of X-rays, not their energy or penetrating power. Concept Tested: X-ray penetrating power: controlled by accelerating voltage (not filament current)

NDA 2011-II

Q. 33. Which one among the following fundamental forces is responsible for the stability of atomic nuclei?

(a)  Gravitational force

(b)  Electromagnetic force

(c)  Strong nuclear force

(d)  Weak nuclear force

Answer: (c) Strong nuclear force

Explanation: The strong nuclear force holds protons and neutrons together in the nucleus: it overcomes the electromagnetic repulsion between protons. It is the strongest of the four fundamental forces but acts only at very short range (~10⁻¹⁵ m). Gravity is far too weak; electromagnetic force would repel protons; the weak force handles radioactive decay. Concept Tested: Strong nuclear force: holds nucleons together in nucleus; overcomes proton-proton repulsion

Q. 34. The atomic number of an element is 17 and its mass number is 35. Which one among the following statements is correct?

(a)  The nucleus contains 17 neutrons and 18 electrons

(b)  The nucleus contains 17 protons and 18 electrons

(c)  The nucleus contains 17 protons and 18 neutrons

(d)  The nucleus contains 18 protons and 17 neutrons

Answer: (c) The nucleus contains 17 protons and 18 neutrons

Explanation: Atomic number Z = 17 → protons = 17. Neutrons = mass number − atomic number = 35 − 17 = 18. The nucleus contains protons and neutrons only: no electrons. Concept Tested: Nuclear composition: protons = 17, neutrons = 35 − 17 = 18; no electrons in nucleus

NDA 2010-II

Q. 35. What does the Sun primarily derive its energy from?

(a)  Nuclear fission

(b)  Nuclear fusion

(c)  Combustion of hydrogen

(d)  Chemical reaction between hydrogen and helium

Answer: (b) Nuclear fusion

Explanation: The Sun generates energy through nuclear fusion: hydrogen nuclei (protons) fuse under extreme temperature and pressure at the core to form helium nuclei. This process converts a tiny mass into enormous energy via E = mc². Fission (splitting) is used in nuclear reactors, not the Sun. Combustion is a chemical process: far too little energy for the Sun’s output. Concept Tested: Sun’s energy: nuclear fusion of hydrogen to helium; E = mc²

Q. 36. Radiation of wavelength 15 nm falls under which region of the electromagnetic spectrum?

(a)  X-ray

(b)  Ultraviolet

(c)  Infrared

(d)  Visible

Answer: (b) Ultraviolet

Explanation: EM spectrum wavelength ranges: X-ray: 0.01–10 nm; UV: 10–380 nm; visible: 380–780 nm; infrared: 780 nm–1 mm. 15 nm falls in the ultraviolet range (just above X-ray boundary at ~10 nm). Concept Tested: EM spectrum: 15 nm falls in ultraviolet (UV: 10–380 nm)

★ JOVIK Exam Insight EM wavelength order (increasing): gamma < X-ray (0.01–10 nm) < UV (10–380 nm) < visible (380–780 nm) < IR < microwave < radio.

NDA 2010-I

Q. 37. The nucleus of ₁₃Al²⁷ contains

(a)  13 protons and 27 neutrons

(b)  13 protons and 14 neutrons

(c)  14 protons and 13 neutrons

(d)  27 protons and 14 neutrons

Answer: (b) 13 protons and 14 neutrons

Explanation: Atomic number Z = 13 → number of protons = 13. Mass number A = 27 → neutrons = A − Z = 27 − 13 = 14. The nucleus contains protons and neutrons only: no electrons. Concept Tested: Nuclear composition: protons = Z, neutrons = A − Z; no electrons in nucleus

Q. 38. In which of the following cases are X-rays produced?

(a)  Alpha rays striking a lead plate

(b)  Fast electrons striking a metal plate

(c)  Gamma rays striking a lead plate

(d)  Slow electrons striking a metal plate

Answer: (b) Fast electrons striking a metal plate

Explanation: X-rays are produced when fast (high-kinetic-energy) electrons decelerate rapidly upon striking a metal (usually tungsten) anode. The kinetic energy is partly converted to X-rays (bremsstrahlung) and characteristic radiation. Slow electrons lack sufficient energy to produce X-rays. Alpha or gamma rays striking a lead plate do not produce X-rays this way. Concept Tested: X-ray production: fast electrons striking metal target (bremsstrahlung)

★ JOVIK Exam Insight X-ray production is one of the most tested concepts: always fast electrons on a metal target. Slow electrons → too little energy. Alpha/gamma on lead → not X-ray production.

Q. 39. What is the ratio of the speed of X-rays to the speed of gamma rays in vacuum?

(a)  Less than one

(b)  Equal to one

(c)  More than one

(d)  Depends upon their frequencies

Answer: (b) Equal to one

Explanation: All electromagnetic waves travel at the same speed in vacuum: c ≈ 3 × 10⁸ m/s. X-rays and gamma rays are both electromagnetic radiation. Their speeds are identical. Speed in vacuum is independent of frequency. Concept Tested: All EM waves travel at same speed in vacuum: ratio of X-ray to gamma ray speed = 1

Q. 40. Among the following types of electromagnetic radiation: (i) X-rays, (ii) Gamma rays: identify which pair is correct

(a)  X-rays and Gamma rays are both electromagnetic

(b)  Alpha rays and Beta rays are both electromagnetic

(c)  Cathode rays and X-rays are both electromagnetic

(d)  Beta rays and Gamma rays are both electromagnetic

Answer: (a) X-rays and Gamma rays are both electromagnetic

Explanation: X-rays and gamma rays are both electromagnetic radiation: they have no charge, travel at speed c, and are part of the EM spectrum. Alpha rays (He²⁺ nuclei), beta rays (electrons), and cathode rays (electron beams) are all particles, not electromagnetic radiation. Concept Tested: Electromagnetic radiation: X-rays and gamma rays; alpha, beta, cathode rays are particles

---