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Dual Nature of Radiation question

2007 · Shift 1 · Q64
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Dual Nature of Radiation question

2007 · Shift 1 · Q64

JEE AdvancedPhysicsDual Nature of RadiationMCQ+3 / −1

Some laws/processes are given in Column I. Match these with the physical phenomena given in Column II and indicate your answer by darkening appropriate bubbles in the 4 ×\times× 4 matrix given in the ORS.

Column I Column II
(A) Transition between two atomic energy levels (P) Characteristic X-rays
(B) Electron emission from a material (Q) Photoelectric effect
(C) Mosley's law (R) Hydrogen spectrum
(D) Change of photon energy into kinetic energy of electrons (S) β\betaβ-decay

  1. A
    A →\to→(P); B →\to→(Q); C →\to→(P); D →\to→ (Q)
  2. B
    A →\to→(R); B →\to→(Q, S); C →\to→(P); D →\to→ (Q, P)
  3. C
    A →\to→(P, R); B →\to→(Q, S); C →\to→(P); D →\to→ (Q)
  4. D
    A →\to→(P, R); B →\to→(Q, S); C →\to→(Q); D →\to→ (P)
View written solutionFree

Correct answer: C

This is a matching-type question where we need to associate the laws/processes in Column I with the corresponding physical phenomena in Column II. Let's analyze each item in Column I one by one.

Step 1: Analyze Column I (A) - Transition between two atomic energy levels

  • A transition of an electron between two discrete energy levels within an atom results in the emission or absorption of a photon. The energy of the photon is equal to the energy difference between the two levels, i.e., Ephoton=ΔE=Einitial−EfinalE_{photon} = \Delta E = E_{initial} - E_{final}Ephoton​=ΔE=Einitial​−Efinal​.
  • (P) Characteristic X-rays: These are produced when an electron from a higher energy level (e.g., L-shell) makes a transition to a lower energy level (e.g., K-shell) to fill a vacancy. This is a transition between two atomic energy levels.
  • (R) Hydrogen spectrum: The spectral lines (like Lyman, Balmer series) in the hydrogen spectrum are produced when an electron in a hydrogen atom de-excites from a higher energy level (n2n_2n2​) to a lower energy level (n1n_1n1​), emitting a photon. This is also a transition between two atomic energy levels.
  • Therefore, (A) matches with both (P) and (R).

Step 2: Analyze Column I (B) - Electron emission from a material

  • This phrase describes processes where electrons are ejected from a substance.
  • (Q) Photoelectric effect: This is the phenomenon where electrons are emitted from a material (usually a metal) when electromagnetic radiation, such as light, hits the material. This perfectly matches the description.
  • (S) β\betaβ-decay: This is a type of radioactive decay in which a beta particle (a fast, energetic electron or positron) is emitted from an atomic nucleus. Since the nucleus is part of the material, this is also a form of electron emission from a material.
  • Other phenomena like thermionic emission also fit this description but are not listed.
  • Therefore, (B) matches with both (Q) and (S).

Step 3: Analyze Column I (C) - Mosley's law

  • Mosley's law is an empirical law concerning the characteristic X-rays emitted by atoms. It states that the square root of the frequency (ν\nuν) of the emitted X-ray is approximately proportional to the atomic number (ZZZ) of the element. The formula is given by ν=a(Z−b)\sqrt{\nu} = a(Z - b)ν​=a(Z−b), where 'a' and 'b' are constants that depend on the type of X-ray line (e.g., Kα\alphaα, Lα\alphaα).
  • (P) Characteristic X-rays: Mosley's law is exclusively related to characteristic X-rays.
  • Therefore, (C) matches only with (P).

Step 4: Analyze Column I (D) - Change of photon energy into kinetic energy of electrons

  • This describes a process where a photon is absorbed and its energy is transferred to an electron, which then possesses kinetic energy.
  • (Q) Photoelectric effect: This is the quintessential example of this process. According to Einstein's photoelectric equation, Kmax=hν−ϕK_{max} = h\nu - \phiKmax​=hν−ϕ, the energy of the incident photon (hνh\nuhν) is converted into the work function (ϕ\phiϕ) to free the electron and the maximum kinetic energy (KmaxK_{max}Kmax​) of the emitted electron.
  • Therefore, (D) matches with (Q).

Step 5: Consolidate the matches and select the correct option

Based on the analysis above, we have the following matches:

  • A →\to→ (P), (R)
  • B →\to→ (Q), (S)
  • C →\to→ (P)
  • D →\to→ (Q)

Now, let's compare these matches with the given options:

  • A: A →\to→(P); B →\to→(Q); C →\to→(P); D →\to→ (Q) - Incorrect because it misses A→\to→R and B→\to→S.
  • B: A →\to→(R); B →\to→(Q, S); C →\to→(P); D →\to→ (Q, P) - Incorrect because it misses A→\to→P and incorrectly includes D→\to→P.
  • C: A →\to→(P, R); B →\to→(Q, S); C →\to→(P); D →\to→ (Q) - This option perfectly matches our derived set of correspondences.
  • D: A →\to→(P, R); B →\to→(Q, S); C →\to→(Q); D →\to→ (P) - Incorrect because C→\to→Q and D→\to→P are wrong.

Thus, the correct option is C.

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