An electron has a de Broglie wavelength of 1 nm. Calculate its momentum and energy

An electron has a de Broglie wavelength of 1 nm. Calculate its momentum and energy.

Answer:-

Given:

• Wavelength (λ) = 1 nm = 1 × 10⁻⁹ m
• Planck’s constant (h) = 6.626 × 10⁻³⁴ J·s
• Electron mass (m) = 9.1 × 10⁻³¹ kg

Step 1: Calculate Momentum (p)

Using the formula:
p = h / λ = 6.626 × 10⁻³⁴ / 1 × 10⁻⁹ = 6.626 × 10⁻²⁵ kg·m/s

Step 2: Calculate Energy (E)

Using the formula:
E = p² / (2m) = (6.626 × 10⁻²⁵)² / (2 × 9.1 × 10⁻³¹) = 4.39 × 10⁻¹⁹ J
Converting to eV:
E ≈ 4.39 × 10⁻¹⁹ / 1.6 × 10⁻¹⁹ ≈ 2.74 eV

Final Answers:

• Momentum (p): 6.626 × 10⁻²⁵ kg·m/s
• Energy (E): 2.74 eV