The de Broglie wavelength \( \lambda \) of a particle is defined by the equation:
\[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE_k}} \]
Here, \( h \) represents Planck's constant, \( p \) is the momentum, \( m \) is the particle's mass, and \( E_k \) denotes its kinetic energy.
When a proton is accelerated by a potential difference \( V \), its kinetic energy is given by:
\[ E_k = eV \]
where \( e \) is the elementary charge.
Substituting \( E_k \) into the de Broglie equation yields:
\[ \lambda = \frac{h}{\sqrt{2meV}} \]
If the accelerating potential is doubled to \( 2V \), the new wavelength \( \lambda' \) is calculated as:
\[ \lambda' = \frac{h}{\sqrt{2m \cdot 2eV}} = \frac{h}{\sqrt{4meV}} \]
This simplifies to:
\[ \lambda' = \frac{h}{2\sqrt{meV}} = \frac{\lambda}{\sqrt{2}} \]
Consequently, the de Broglie wavelength diminishes when the accelerating potential is doubled. Hence, the correct response is "Decreases".