Aproton accelerated through a potential difference V has a de Broglie wavelength λ. Ondoubling the accelerating potential, de Broglie wavelength of the proton _________.

The correct option is decreases.

To understand why this is the correct answer, let us look at the relationship between the de Broglie wavelength of a charged particle and its accelerating potential difference.

The de Broglie wavelength ($\lambda$) of a particle is given by the formula:
$\lambda =\frac{h}{p}$
where $h$ is Planck's constant and $p$ is the momentum of the particle.

The momentum $p$ of a particle of mass $m$ can be related to its kinetic energy ($K$) as follows:
$p=\sqrt{2mK}$

When a particle with charge $q$ (such as a proton) is accelerated from rest through a potential difference $V$, the work done on it by the electric field becomes its kinetic energy:
$K=qV$

Substituting this expression for kinetic energy into the momentum equation gives:
$p=\sqrt{2mqV}$

Now, substituting the expression for momentum back into the de Broglie wavelength equation, we get:
$\lambda =\frac{h}{\sqrt{2mqV}}$

For a given particle (like a proton), the Planck's constant ($h$), mass ($m$), and charge ($q$) are constants. Therefore, the de Broglie wavelength is inversely proportional to the square root of the accelerating potential difference ($V$):
$\lambda \propto \frac{1}{\sqrt{V}}$

Since the accelerating potential difference $V$ is doubled (increased to $2V$), the new wavelength ${\lambda }^{\prime }$ will be:
${\lambda }^{\prime }=\frac{\lambda }{\sqrt{2}}\approx 0.707\lambda$

Because the new wavelength is smaller than the original wavelength, the de Broglie wavelength decreases.