Preliminary Models and Dual Nature of Light – Nature of electromagnetic radiation: NEET MCQs with Answers

15 free questions on Preliminary Models and Dual Nature of Light – Nature of electromagnetic radiation for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Nature of electromagnetic radiation, photoelectric effect;

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Question 1

Previous-year question
An isotope of an element has 19.23% more neutrons than protons. Identify the element along with its mass number. (Given: atomic number Fe = 26, Co = 27) [NEET (Sawai Madhopur) 2024]
  1. ⁵⁷Fe
  2. ⁵⁷Co
  3. ⁶⁰Co
  4. ⁵⁶Fe
Show answer
Answer: (A) ⁵⁷Fe
Explanation: Neutrons = 1.1923 × protons. For Fe (Z = 26): n = 1.1923 × 26 = 31, so A = 26 + 31 = 57, i.e. ⁵⁷Fe. Check others: ⁵⁷Co has 30 neutrons (30/27 = 1.111), ⁶⁰Co has 33 (1.222), ⁵⁶Fe has 30 (1.154) — none gives 19.23% excess.

Question 2

Previous-year question
An infrared range finder with a power of 1 mW emits monochromatic photons of wavelength 1000 nm for 0.1 second. How many photons are emitted in this 0.1 second? (Given: h = 6.626 × 10⁻³⁴ J s, c = 3 × 10⁸ m s⁻¹, Avogadro number = 6.022 × 10²³) [NEET (Overseas) 2022]
  1. 5 × 10¹⁴
  2. 30 × 10³⁴
  3. 5 × 10¹¹
  4. 30 × 10³⁷
Show answer
Answer: (A) 5 × 10¹⁴
Explanation: Energy emitted = power × time = 1 × 10⁻³ W × 0.1 s = 1 × 10⁻⁴ J. Energy of one photon E = hc/λ = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(1000 × 10⁻⁹) = 1.988 × 10⁻¹⁹ J. Number of photons = 1 × 10⁻⁴/1.988 × 10⁻¹⁹ ≈ 5 × 10¹⁴.

Question 3

Previous-year question
Calculate the energy in joule corresponding to light of wavelength 45 nm. (Planck's constant h = 6.63 × 10⁻³⁴ J s; speed of light c = 3 × 10⁸ m s⁻¹) [AIPMT 2014]
  1. 6.67 × 10¹⁵
  2. 6.67 × 10¹¹
  3. 4.42 × 10⁻¹⁵
  4. 4.42 × 10⁻¹⁸
Show answer
Answer: (D) 4.42 × 10⁻¹⁸
Explanation: E = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(45 × 10⁻⁹) = 1.989 × 10⁻²⁵/4.5 × 10⁻⁸ = 4.42 × 10⁻¹⁸ J.

Question 4

Previous-year question
In photoelectron emission, the energy of the emitted electron is [AIPMT 1994]
  1. greater than the incident photon
  2. same as that of the incident photon
  3. smaller than the incident photon
  4. proportional to the intensity of incident photon
Show answer
Answer: (C) smaller than the incident photon
Explanation: By Einstein's equation, KE = hν − hν₀: part of the photon's energy is used up as the work function, so the emitted electron always has less energy than the incident photon. KE depends on frequency, while intensity decides only the number of electrons emitted.

Question 5

Previous-year question
Which of the following is never true for cathode rays? [AIPMT 1994]
  1. They possess kinetic energy
  2. They are electromagnetic waves
  3. They produce heat
  4. They produce mechanical pressure
Show answer
Answer: (B) They are electromagnetic waves
Explanation: Cathode rays are streams of electrons, i.e. material particles with mass and charge. They therefore possess kinetic energy, heat the target and exert mechanical pressure (they rotate a paddle wheel), and they are deflected by electric and magnetic fields, so they are never electromagnetic waves.

Question 6

Previous-year question
The energy of a photon is given as ΔE/atom = 3.03 × 10⁻¹⁹ J atom⁻¹. Then the wavelength (λ) of the photon is (Given h = 6.63 × 10⁻³⁴ Js, c = 3.00 × 10⁸ ms⁻¹) [AIPMT 2000]
  1. 6.56 nm
  2. 65.6 nm
  3. 656 nm
  4. 0.656 nm
Show answer
Answer: (C) 656 nm
Explanation: From E = hc/λ, rearranging gives λ = hc/E. λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸)/(3.03 × 10⁻¹⁹) = (1.989 × 10⁻²⁵)/(3.03 × 10⁻¹⁹) = 6.56 × 10⁻⁷ m. Converting: 6.56 × 10⁻⁷ m = 656 × 10⁻⁹ m = 656 nm. Worth remembering: 656 nm is the red Hα line of the Balmer series of hydrogen.

Question 7

Previous-year question
The value of Planck’s constant is 6.63 × 10⁻³⁴ Js and the velocity of light is 3.0 × 10⁸ ms⁻¹. Which value is closest to the wavelength in nanometres of a quantum of light with frequency 8 × 10¹⁵ s⁻¹? [AIPMT 2003]
  1. 4 × 10¹
  2. 3 × 10⁷
  3. 2 × 10⁻²⁵
  4. 5 × 10⁻¹⁸
Show answer
Answer: (A) 4 × 10¹
Explanation: λ = c/ν = (3.0 × 10⁸ m s⁻¹)/(8 × 10¹⁵ s⁻¹) = 0.375 × 10⁻⁷ m = 3.75 × 10⁻⁸ m. Converting to nanometres (1 m = 10⁹ nm): 3.75 × 10⁻⁸ m = 37.5 nm.
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Question 8

Previous-year question
The energy absorbed by each molecule (A₂) of a substance is 4.4 × 10⁻¹⁹ J and the bond energy per molecule is 4.0 × 10⁻¹⁹ J. The kinetic energy of the molecule per atom will be [AIPMT 2009]
  1. 2.0 × 10⁻²⁰ J
  2. 2.2 × 10⁻¹⁹ J
  3. 2.0 × 10⁻¹⁹ J
  4. 4.0 × 10⁻²⁰ J
Show answer
Answer: (A) 2.0 × 10⁻²⁰ J
Explanation: KE released per molecule = 4.4 × 10⁻¹⁹ − 4.0 × 10⁻¹⁹ = 0.4 × 10⁻¹⁹ J. This energy is shared equally by the two A atoms, so KE per atom = 0.2 × 10⁻¹⁹ = 2.0 × 10⁻²⁰ J.

Question 9

Previous-year question
The energies E₁ and E₂ of two radiations are 25 eV and 50 eV respectively. The relation between their wavelengths, i.e. λ₁ and λ₂, will be [AIPMT 2011]
  1. λ₁ = 2λ₂
  2. λ₁ = 4λ₂
  3. λ₁ = ½ λ₂
  4. λ₁ = λ₂
Show answer
Answer: (A) λ₁ = 2λ₂
Explanation: Since E = hc/λ, energy and wavelength are inversely proportional: E ∝ 1/λ. Therefore E₁/E₂ = λ₂/λ₁. 25/50 = λ₂/λ₁ ⇒ λ₂/λ₁ = 1/2 ⇒ λ₁ = 2λ₂. Sense-check: the radiation of LOWER energy (25 eV) must have the LONGER wavelength, which is exactly what λ₁ = 2λ₂ states.

Question 10

Previous-year question
The value of Planck’s constant is 6.63 × 10⁻³⁴ Js. The speed of light is 3 × 10¹⁷ nm s⁻¹. Which value is closest to the wavelength in nanometres of a quantum of light with frequency 6 × 10¹⁵ s⁻¹? [NEET 2013]
  1. 10
  2. 25
  3. 50
  4. 75
Show answer
Answer: (C) 50
Explanation: The relation between speed, frequency and wavelength is c = νλ, so λ = c/ν. Note that c has been supplied directly in nm s⁻¹, so the answer comes out in nm with no conversion needed. λ = (3 × 10¹⁷ nm s⁻¹)/(6 × 10¹⁵ s⁻¹) = 0.5 × 10² = 50 nm. Planck’s constant is not required at all here — it is supplied only as a distractor.

Question 11

Previous-year question
Calculate the energy in joule corresponding to light of wavelength 45 nm. (Planck’s constant h = 6.63 × 10⁻³⁴ Js; speed of light c = 3 × 10⁸ ms⁻¹) [AIPMT 2014]
  1. 6.67 × 10¹⁵
  2. 6.67 × 10¹¹
  3. 4.42 × 10⁻¹⁵
  4. 4.42 × 10⁻¹⁸
Show answer
Answer: (D) 4.42 × 10⁻¹⁸
Explanation: E = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(45 × 10⁻⁹) = 1.989 × 10⁻²⁵/4.5 × 10⁻⁸ = 4.42 × 10⁻¹⁸ J.

Question 12

Previous-year question
The number of protons, neutrons and electrons in ₁₇₁⁵Lu (mass number 175, atomic number 71) respectively are [NEET 2020]
  1. 71, 104 and 71
  2. 104, 71 and 71
  3. 71, 71 and 104
  4. 175, 104 and 71
Show answer
Answer: (A) 71, 104 and 71
Explanation: Atomic number Z = 71 ⇒ number of protons = 71. For a neutral atom, number of electrons = number of protons = 71. Number of neutrons = mass number − atomic number = 175 − 71 = 104. Hence protons : neutrons : electrons = 71 : 104 : 71.

Question 13

Which of the following electromagnetic radiations has the highest frequency?
  1. Red light
  2. Infrared radiation
  3. Microwaves
  4. X-rays
Show answer
Answer: (D) X-rays
Explanation: Frequency is inversely proportional to wavelength because c = νλ. Of these radiations, X-rays have the shortest wavelength and so the highest frequency. In order of increasing frequency they are microwaves, infrared, red light and X-rays.

Question 14

The charge-to-mass ratio of the electron was first measured by studying how cathode rays are deflected by electric and magnetic fields. This measurement was made by:
  1. E. Rutherford
  2. R.A. Millikan
  3. J.J. Thomson
  4. J. Chadwick
Show answer
Answer: (C) J.J. Thomson
Explanation: J.J. Thomson measured the ratio e/mₑ in 1897 from the deflection of cathode rays in electric and magnetic fields. Millikan later found the charge of the electron with his oil-drop experiment, Rutherford proposed the nuclear model and Chadwick discovered the neutron.

Question 15

Which of the following correctly orders the electron, the proton and the neutron by increasing mass?
  1. Electron < neutron < proton
  2. Electron < proton < neutron
  3. Neutron < proton < electron
  4. Proton < electron < neutron
Show answer
Answer: (B) Electron < proton < neutron
Explanation: The electron has a mass of about 9.11 × 10⁻³¹ kg, far less than the nucleons. The proton has a mass of about 1.6726 × 10⁻²⁷ kg and the neutron about 1.6749 × 10⁻²⁷ kg, so the neutron is slightly heavier than the proton.

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