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JEE Main 2020 (5 Sep, Shift 1) · Davisson-Germer, Uncertainty, Compton & X-Rays

Davisson-Germer, Uncertainty, Compton & X-Rays question from JEE Main 2020 (5 Sep, Shift 1)

Q10JEE Main 2020 · 5 SEP · SHIFT 1Difficult

Davisson-Germer, Uncertainty, Compton & X-Rays

A beam of electrons of energy E scatters from a target having atomic spacing of 1 A˚\text{\AA}. The first maximum intensity occurs at θ\theta = 60o. Then E (in eV) is ______.

(Planck constant h = 6.64 × 10–34 Js,

1 eV = 1.6 × 10–19 J, electron

mass m = 9.1 × 10–31 kg)

Numerical answer type. Work it out, then open the answer.

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Answer: 50.47

  1. Bragg condition for first maximum: 2dsin⁡θ=λ2d\sin\theta = \lambda
  2. λ=2×1×sin⁡60∘=3 A˚=1.732×10−10 m\lambda = 2 \times 1 \times \sin 60^\circ = \sqrt{3}\text{ \AA} = 1.732 \times 10^{-10}\text{ m}
  3. E=h22mλ2=(6.64×10−34)22×9.1×10−31×3×10−20=8.07×10−18 JE = \frac{h^2}{2m\lambda^2} = \frac{(6.64 \times 10^{-34})^2}{2 \times 9.1 \times 10^{-31} \times 3 \times 10^{-20}} = 8.07 \times 10^{-18}\text{ J}
  4. E=8.07×10−181.6×10−19≈50.47 eVE = \frac{8.07 \times 10^{-18}}{1.6 \times 10^{-19}} \approx 50.47\text{ eV}
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