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NEET 2012 (AIPMT, Prelims) · Matter Waves: de Broglie Wavelength

Matter Waves: de Broglie Wavelength question from NEET 2012 (AIPMT, Prelims)

Q143NEET 2012 · AIPMT · PRELIMSMedium

Matter Waves: de Broglie Wavelength

An α\alpha-particle moves in a circular path of radius 0.83 cm in the presence of a magnetic field of 0.25 Wb/m2. The de Broglie wavelength associated with the particle will be

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Answer: (d)

  1. Circular motion: r=mvqB⇒p=qBrr = \dfrac{mv}{qB} \Rightarrow p = qBr, so λ=hqBr\lambda = \dfrac{h}{qBr}.
  2. q=2e=3.2×10−19q = 2e = 3.2\times10^{-19} C, qBr=3.2×10−19×0.25×0.83×10−2=6.64×10−22qBr = 3.2\times10^{-19}\times0.25\times0.83\times10^{-2} = 6.64\times10^{-22}
  3. λ=6.63×10−346.64×10−22≈10−12\lambda = \dfrac{6.63\times10^{-34}}{6.64\times10^{-22}} \approx 10^{-12} m =0.01= 0.01 Å
  4. So the answer is (d) 0.01 Å.
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