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Physics · Atomic Physics

JEE Main 2026 — 23 January, Morning Shift — Question 30

The de Broglie wavelength of an oxygen molecule at 27∘C27^{\circ} \mathrm{C} is x×10−12 m\mathrm{x} \times 10^{-12} \mathrm{~m}. The value of x is (take Planck's constant =6.63×10−34 J.s=6.63 \times 10^{-34} \mathrm{~J} . \mathrm{s}, Boltzmann constant =1.38×10−23 J/K=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}, mass of oxygen. Molecule =5.31×10−26 kg=5.31 \times 10^{-26} \mathrm{~kg} ).

  1. Option A:

    26

    Correct
  2. Option B:

    24

  3. Option C:

    30

  4. Option D:

    20

Answer: A

Step-by-step solution

λ=h2mK=h2 m(32kT)\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}\left(\frac{3}{2} \mathrm{kT}\right)}} λ=h3mkT\lambda=\frac{\mathrm{h}}{\sqrt{3 \mathrm{mkT}}} =6.63×10−343×5.31×10−26×1.38×10−23×300=\frac{6.63 \times 10^{-34}}{\sqrt{3 \times 5.31 \times 10^{-26} \times 1.38 \times 10^{-23} \times 300}} =2.58×10−11=25.8×10−12=2.58 \times 10^{-11}=25.8 \times 10^{-12} So, x=26x=26

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
23 January, Morning Shift
Subject
Physics
Chapter
Atomic Physics
Topic
Dual Nature of Matter

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The de Broglie wavelength of an oxygen molecule at 27 ° C is x × 10… | JEE Main 2026 PYQ with Solution · DhiX AI