To solve this problem, we need to understand the relationship between kinetic energy and the de Broglie wavelength.$\\$The de Broglie wavelength $(\lambda)$ of a particle is given by:$\\ \lambda = \frac{h}{p} \\$where $h$ is Planck's constant and $p$ is the momentum of the particle.$\\$The momentum $(p)$ can be expressed in terms of kinetic energy $(K)$ as:$\\ p = \sqrt{2mK} \\$where $m$ is the mass of the particle.$\\$Substituting this into the expression for $\lambda: \\ \lambda = \frac{h}{\sqrt{2mK}} \\$If the kinetic energy is increased to 16 times its original value, the new kinetic energy $(K')$ is:$\\ K' = 16K \\$The new de Broglie wavelength $(\lambda')$ is:$\\ \lambda' = \frac{h}{\sqrt{2mK'}} = \frac{h}{\sqrt{2m(16K)}} \\$Simplifying:$\\ \lambda' = \frac{h}{\sqrt{32mK}} = \frac{h}{4\sqrt{2mK}} \\$Thus, $\lambda' = \frac{\lambda}{4} \\$The percentage change in the de Broglie wavelength is calculated as:$\\$Percentage change$= \left( \frac{\lambda' - \lambda}{\lambda} \right) \times 100\% \\$Substitute $\lambda' = \frac{\lambda}{4}: \\$Percentage change$= \left( \frac{\frac{\lambda}{4} - \lambda}{\lambda} \right) \times 100\% \\$Simplify:$\\$Percentage change$= \left( \frac{\lambda - 4\lambda}{4\lambda} \right) \times 100\% \\$Percentage change$= \left( \frac{-3\lambda}{4\lambda} \right) \times 100\% \\$Percentage change$= -75\% \\$Therefore, the de Broglie wavelength decreases by $75\%$, which corresponds to Option 2.