If the kinetic energy of an electron is 100 eV, what is the wavelength of the de-Broglie wave associated with…

Analyze the question: We need the de-Broglie wavelength of an electron whose kinetic energy is 100 eV.

Concept: The de-Broglie wavelength is λ = h/p, where p is the momentum. Since KE = p²/2m, we get p = √(2mKE), so

λ = h / √(2 m KE)

Solution:

1. Convert the energy to joules: KE = 100 × 1.6 × 10⁻¹⁹ = 1.6 × 10⁻¹⁷ J

2. Find 2m·KE: 2 × 9.1 × 10⁻³¹ × 1.6 × 10⁻¹⁷ = 29.12 × 10⁻⁴⁸ = 2.912 × 10⁻⁴⁷

3. Take the square root: √(2.912 × 10⁻⁴⁷) = √(29.12 × 10⁻⁴⁸) = 5.396 × 10⁻²⁴ kg m/s

4. Find the wavelength: λ = 6.6 × 10⁻³⁴ / 5.396 × 10⁻²⁴ ≈ 1.22 × 10⁻¹⁰ m

Answer: The wavelength is about 1.22 × 10⁻¹⁰ m, which is option A.

Tip: For electrons, a handy shortcut is λ ≈ 1.226/√V nm, where V is the accelerating voltage (or the energy in eV). For 100 eV: 1.226/10 = 0.1226 nm = 1.22 × 10⁻¹⁰ m. Also, remember to convert eV to joules before using the formula, as forgetting this is the most common mistake.