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JEE Main 2019 (10 Jan, Shift 1) · Davisson-Germer, Uncertainty, Compton & X-Rays

Davisson-Germer, Uncertainty, Compton & X-Rays question from JEE Main 2019 (10 Jan, Shift 1)

Q11JEE Main 2019 · 10 JAN · SHIFT 1Medium

Davisson-Germer, Uncertainty, Compton & X-Rays

In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of 7.5 × 10–12 m, the minimum electron energy required is close to -

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Answer: (a)

  1. Required wavelength λ=7.5×10−12 m\lambda = 7.5 \times 10^{-12}\text{ m}, and λ=hp\lambda = \frac{h}{p}, E=p22mE = \frac{p^2}{2m}.
  2. p=6.63×10−347.5×10−12=8.84×10−23 kg m/sp = \frac{6.63 \times 10^{-34}}{7.5 \times 10^{-12}} = 8.84 \times 10^{-23}\text{ kg m/s}
  3. E=(8.84×10−23)22×9.1×10−31≈4.3×10−15 JE = \frac{(8.84 \times 10^{-23})^2}{2 \times 9.1 \times 10^{-31}} \approx 4.3 \times 10^{-15}\text{ J}
  4. E=4.3×10−151.6×10−19≈2.7×104 eV≈25 keVE = \frac{4.3 \times 10^{-15}}{1.6 \times 10^{-19}} \approx 2.7 \times 10^{4}\text{ eV} \approx 25\text{ keV}
  5. So the answer is (a) 25 keV.
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