If the kinetic energy of a particle is increased by 16 times, the percentage change in the de Broglie wavelength of the particle is:

Let the initial de Broglie wavelength be λ₀ = h/mv₀
The kinetic energy is increased 16 times.
Thus KE′ = 16KE₀ ⇒ 1/2mv² = 16 × 1/2mv₀² ⇒ v = 4v₀
Thus the new de Broglie wavelength = λ = h/mv = h/m(4v₀) = λ₀/4
Thus the percentage change in wavelength = (λ₀ - λ)/λ₀ × 100 = 75