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JEE Main 2019 (11 Jan, Shift 1) · Matter Waves: de Broglie Wavelength

Matter Waves: de Broglie Wavelength question from JEE Main 2019 (11 Jan, Shift 1)

Q123JEE Main 2019 · 11 JAN · SHIFT 1Medium

Matter Waves: de Broglie Wavelength

If the de Broglie wavelength of an electron is equal to 10–3 times the wavelength of a photon of frequency 6 ×\times 1014 Hz, then the speed of electron is equal to: (Speed of light = 3 ×\times 108 m/s, Planck's constant = 6.63 ×\times 10–34 J.s, Mass of electron = 9.1 ×\times 10–31 kg)

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Answer: (b)

  1. Photon: λp=cf=3×1086×1014=5×10−7\lambda_p = \dfrac{c}{f} = \dfrac{3\times10^8}{6\times10^{14}} = 5\times10^{-7} m.
  2. Electron: λe=10−3λp=5×10−10\lambda_e = 10^{-3}\lambda_p = 5\times10^{-10} m.
  3. v=hmλe=6.63×10−349.1×10−31×5×10−10≈1.45×106v = \dfrac{h}{m\lambda_e} = \dfrac{6.63\times10^{-34}}{9.1\times10^{-31}\times5\times10^{-10}} \approx 1.45\times10^6 m/s
  4. So the answer is (b).
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