MCQ
An electron of mass $m$ when accelerated through a potential difference $V$ has de-Broglie wavelength $\lambda$. The de-Broglie wavelength associated with a proton of mass $M$ accelerated throug the same potential difference will be
- A(a) $\lambda \frac{m}{M}$
- ✓(b) $\lambda \sqrt{\frac{m}{M}}$
- C(c) $\lambda \frac{M}{m}$
- D(d) $\lambda \sqrt{\frac{M}{m}}$

