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JEE Main 2022 (30 Jul, Shift 2) · Matter Waves: de Broglie Wavelength

Matter Waves: de Broglie Wavelength question from JEE Main 2022 (30 Jul, Shift 2)

Q81JEE Main 2022 · 30 JUL · SHIFT 2Easy

Matter Waves: de Broglie Wavelength

The de Broglie wavelength associated with an electron, accelerated through a potential difference of 15056 V15056\text{ V}, is:

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Answer: (a)

  1. For an electron, λ=12.27V\lambda = \dfrac{12.27}{\sqrt{V}} Å.
  2. 15056≈122.7\sqrt{15056} \approx 122.7, so λ=12.27122.7=0.1\lambda = \dfrac{12.27}{122.7} = 0.1 Å.
  3. So the answer is (a) 0.1 Å.
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