A beam of electrons of energy E scatters from a target having atomic spacing of 1 Aᵒ. The first maximum intensity occurs at θ = 60ᵒ. Then E (in eV) is ______. (Round off to the nearest integer)
(Planck constant h = 6.64 × 10 –34 Js,
1 eV = 1.6 × 10 –19 J, electron
mass m = 9.1 × 10 –31 kg)

Solution

Physics sir · 4 steps

Bragg's condition gives you the electron's wavelength, and de Broglie turns that wavelength into energy.

1. Find the wavelength. Take the 1 Å spacing as the gap d between atomic planes and 60° as the glancing angle. The first maximum is n = 1 in Bragg's law 2d sin θ = nλ.

   λ = 2d sin θ = 2 × (1 Å) × √3/2 = √3 Å

2. Wavelength to energy. For an electron, λ = h/√2mE, so E = h²/2mλ². Here λ² = 3 × 10⁻²⁰ m².
3. Put in numbers. h² = (6.64 × 10⁻³⁴)² = 4.409 × 10⁻⁶⁷ and 2mλ² = 2 × 9.1 × 10⁻³¹ × 3 × 10⁻²⁰ = 5.46 × 10⁻⁵⁰. So E = (4.409 × 10⁻⁶⁷)/(5.46 × 10⁻⁵⁰) = 8.075 × 10⁻¹⁸ J.
4. So. E = (8.075 × 10⁻¹⁸)/(1.6 × 10⁻¹⁹) ≈ 50.5 eV, which rounds to 50.

The common slip: 202. Using λ = d sin θ instead of Bragg's 2d sin θ = λ halves the wavelength and makes E four times larger, giving about 202 instead of 50.

The idea to keep. For electron diffraction, get λ from the diffraction condition first, then use E = h²/2mλ². Since E ∝ 1/λ², any factor of 2 in λ changes E by 4.

Answer from the NTA final answer key. Solution by PrepWiser.