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JEE Main 2026 (23 Jan, Shift 1) · Matter Waves: de Broglie Wavelength

Matter Waves: de Broglie Wavelength question from JEE Main 2026 (23 Jan, Shift 1)

Q15JEE Main 2026 · 23 JAN · SHIFT 1Medium

Matter Waves: de Broglie Wavelength

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

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Answer: (d)

  1. λ=h3mkT\lambda = \dfrac{h}{\sqrt{3mkT}} with T=300 KT = 300\text{ K}.
  2. 3mkT=3×5.31×10−26×1.38×10−23×300≈6.60×10−463mkT = 3 \times 5.31 \times 10^{-26} \times 1.38 \times 10^{-23} \times 300 \approx 6.60 \times 10^{-46}, so 3mkT≈2.57×10−23\sqrt{3mkT} \approx 2.57 \times 10^{-23}
  3. λ=6.63×10−342.57×10−23≈2.58×10−11 m≈26×10−12 m\lambda = \dfrac{6.63 \times 10^{-34}}{2.57 \times 10^{-23}} \approx 2.58 \times 10^{-11}\text{ m} \approx 26 \times 10^{-12}\text{ m}
  4. So the answer is (d) 26.
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