A positron (m) and a proton (M) are accelerated by the same accelerating potential. Then the ratio of the associated wavelength of positron and proton will be

The charge of positron and proton are same and they are accelerated through the same potential difference, therefore kinetic energy acquired by them will be the same.

$K.E. = \mathrm{eV}$

The relation between de Broglie wavelength and mass and energy of a particle is given by

$\lambda = \frac{h}{\sqrt{2m(K.E.)}}$

${\lambda }_{positron} = \frac{h}{\sqrt{2m_{positron}(K.E.)}}$

${\lambda }_{proton} = \frac{h}{\sqrt{2m_{proton}(K.E.)}}$

$\frac{{\lambda }_{positron}}{{\lambda }_{proton}} = \sqrt{\frac{m_{proton}}{m_{positron}}} = \sqrt{\frac{M}{m}}$