A particle having electric charge 3 × 10⁻¹⁹ C and mass 6 × 10⁻²⁷ kg is accelerated by applying an electric potential of 1.21 V. Wavelength of the matter wave associated with the particle is α × 10⁻¹² m. The value of α i $s _________.$
(Take Planck’s constant = 6.6 × 10⁻³⁴ J.s)

Question

A particle having electric charge 3 × 10⁻¹⁹ C and mass 6 × 10⁻²⁷ kg is accelerated by applying an electric potential of 1.21 V. Wavelength of the matter wave associated with the particle is α × 10⁻¹² m. The value of α i $s _________.$

(Take Planck’s constant = 6.6 × 10⁻³⁴ J.s)

G1 Physics · Accepted Answer

A charged particle accelerated from rest through potential V gains kinetic energy qV = p²/(2m). Therefore, p = √(2mqV). The de Broglie wavelength is λ = h/p = h/√(2mqV). Substitute: h = 6.6 × 10⁻³⁴ J·s, m = 6 × 10⁻²⁷ kg, q = 3 × 10⁻¹⁹ C, V = 1.21 V. Thus, λ = (6.6 × 10⁻³⁴)/√[2 × 6 × 10⁻²⁷ × 3 × 10⁻¹⁹ × 1.21]. The quantity under the square root is 43.56 × 10⁻⁴⁶. Its square root is 6.6 × 10⁻²³. Hence, λ = (6.6 × 10⁻³⁴)/(6.6 × 10⁻²³) = 10⁻¹¹ m = 10 × 10⁻¹² m. Therefore, α = 10.