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NEET 2019 · Matter Waves: de Broglie Wavelength

Matter Waves: de Broglie Wavelength question from NEET 2019

Q120NEET 2019Easy

Matter Waves: de Broglie Wavelength

An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is (nearly): (me = 9 ×\times 10–31 kg)

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Answer: (a)

  1. λ=1.227V\lambda = \dfrac{1.227}{\sqrt{V}} nm =1.227104= \dfrac{1.227}{\sqrt{10^4}} nm =0.01227= 0.01227 nm.
  2. λ≈12.2×10−12\lambda \approx 12.2\times10^{-12} m.
  3. So the answer is (a) 12.2×10−1212.2\times10^{-12} m.
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