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NEET 1996 (AIPMT) · Matter Waves: de Broglie Wavelength

Matter Waves: de Broglie Wavelength question from NEET 1996 (AIPMT)

Q148NEET 1996 · AIPMTEasy

Matter Waves: de Broglie Wavelength

An electron beam has a kinetic energy equal to 100 eV. Find its wavelength associated with a beam, if mass of electron = 9.1×10−319.1 \times 10^{-31} kg and 1 eV = 1.6×10−191.6 \times 10^{-19} J/eV.

(Planck's constant = 6.6×10−346.6 \times 10^{-34} Js)

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Answer: (c)

  1. λ=h2mK\lambda = \dfrac{h}{\sqrt{2mK}}, with K=100×1.6×10−19=1.6×10−17K = 100\times1.6\times10^{-19} = 1.6\times10^{-17} J.
  2. 2×9.1×10−31×1.6×10−17≈5.4×10−24\sqrt{2\times9.1\times10^{-31}\times1.6\times10^{-17}} \approx 5.4\times10^{-24} kg m/s.
  3. λ=6.6×10−345.4×10−24≈1.2×10−10\lambda = \dfrac{6.6\times10^{-34}}{5.4\times10^{-24}} \approx 1.2\times10^{-10} m.
  4. So the answer is (c) 1.2 Å.
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