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JEE Main 2013 (22 Apr, Shift 1) · Matter Waves: de Broglie Wavelength

Matter Waves: de Broglie Wavelength question from JEE Main 2013 (22 Apr, Shift 1)

Q141JEE Main 2013 · 22 APR · SHIFT 1Difficult

Matter Waves: de Broglie Wavelength

Orbits of a particle moving in a circle are such that the perimeter of the orbit equals an integer number of de-Broglie wavelengths of the particle. For a charged particle moving in a plane perpendicular to a magnetic field, the radius of the nthn^{th} orbital will therefore be proportional to:

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Answer: (c)

  1. Condition: 2πr=nλ=nhp2\pi r = n\lambda = \dfrac{nh}{p}.
  2. In field BB: p=qBrp = qBr.
  3. 2πr=nhqBr⇒r2=nh2πqB⇒r∝n1/22\pi r = \dfrac{nh}{qBr} \Rightarrow r^2 = \dfrac{nh}{2\pi qB} \Rightarrow r \propto n^{1/2}
  4. So the answer is (c).
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