A proton accelerated through a potential difference V has a de Broglie wavelength $\lambda$. On doubling the accelerating potential, de Broglie wavelength of the proton ________.

Fill in the blank with the correct answer from the options given below.

decreases

$\lambda =\frac{h}{p} = \frac{h}{\sqrt{2mqV}}$

On doubling the accelerating potential the wavelength becomes $\frac{1}{\sqrt{2}}$ times.

The correct answer is Option (4) → decreases