An electron of mass $m$ has de-Broglie wavelength $\lambda$ when accelerated through potential difference $V$. When a proton of mass $M$ is accelerated through a potential difference of $9V$,the de-Broglie wavelength associated with it will be (Assume that wavelength is determined at low voltage).

• A
  $\frac{\lambda}{3} \sqrt{\frac{M}{m}}$
• B
  $\frac{\lambda}{3} \cdot \frac{M}{m}$
• C
  $\frac{\lambda}{3} \sqrt{\frac{m}{M}}$
• D
  $\frac{\lambda}{3} \cdot \frac{m}{M}$