Correct Option: (D) **Step-by-step Solution:** 1. **Recall the de Broglie wavelength formula:** The de Broglie wavelength ($\lambda$) of a particle is inversely proportional to its momentum ($p$): $$\lambda = \frac{h}{p} \quad \text{(Equation 1)}$$ where $h$ is Planck's constant. 2. **Relate momentum to kinetic energy:** The kinetic energy ($E$) of a particle with mass $m$ and momentum $p$ is given by: $$E = \frac{p^2}{2m}$$ From this, we can express momentum in terms of kinetic energy: $$p^2 = 2mE$$ $$p = \sqrt{2mE} \quad \text{(Equation 2)}$$ 3. **Express de Broglie wavelength in terms of kinetic energy:** Substitute Equation 2 into Equation 1: $$\lambda = \frac{h}{\sqrt{2mE}} \quad \text{(Equation 3)}$$ 4. **Consider the case when kinetic energy is doubled:** Let the initial kinetic energy be $E_1 = E$, and the initial de Broglie wavelength be $\lambda_1 = \lambda$. So, from Equation 3: $$\lambda = \frac{h}{\sqrt{2mE}} \quad \text{(Initial state)}$$ Now, the kinetic energy is doubled, so the new kinetic energy $E_2 = 2E$. Let the new de Broglie wavelength be $\lambda_2$. Applying Equation 3 for the new state: $$\lambda_2 = \frac{h}{\sqrt{2m(2E)}}$$ $$\lambda_2 = \frac{h}{\sqrt{4mE}}$$ $$\lambda_2 = \frac{h}{2\sqrt{mE}}$$ 5. **Calculate the new wavelength in terms of the original wavelength:** We can rewrite the expression for $\lambda_2$: $$\lambda_2 = \frac{1}{\sqrt{2}} \left( \frac{h}{\sqrt{2mE}} \right)$$ Recognizing that $\frac{h}{\sqrt{2mE}}$ is the original wavelength $\lambda$: $$\lambda_2 = \frac{\lambda}{\sqrt{2}}$$ Therefore, if the kinetic energy of the particle is doubled, its de Broglie wavelength will decrease by a factor of $\sqrt{2}$. The final answer is $\boxed{\text{(D)}}$