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JEE Main 2021 (1 Sep, Shift 2) · Matter Waves: de Broglie Wavelength

Matter Waves: de Broglie Wavelength question from JEE Main 2021 (1 Sep, Shift 2)

Q102JEE Main 2021 · 1 SEP · SHIFT 2Medium

Matter Waves: de Broglie Wavelength

The temperature of an ideal gas in 3-dimensions is 300 K. The corresponding de-Broglie wavelength of the electron approximately at 300 K, is:

[me = mass of electron = 9 ×\times 10−-31 kg, h = Planck constant = 6.6 ×\times 10−-34 Js, kB = Boltzmann constant = 1.38 ×\times 10−-23 JK−1{}^{-1}]

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Answer: (a)

  1. λ=h3mkT\lambda = \dfrac{h}{\sqrt{3mkT}} (using K=32kTK = \frac{3}{2}kT).
  2. 3mkT=3×9×10−31×1.38×10−23×300≈1.118×10−503mkT = 3 \times 9 \times 10^{-31} \times 1.38 \times 10^{-23} \times 300 \approx 1.118 \times 10^{-50}, so 3mkT≈1.057×10−25\sqrt{3mkT} \approx 1.057 \times 10^{-25}
  3. λ=6.6×10−341.057×10−25≈6.2×10−9 m\lambda = \dfrac{6.6 \times 10^{-34}}{1.057 \times 10^{-25}} \approx 6.2 \times 10^{-9}\text{ m}
  4. So the answer is (a) 6.26 nm.
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