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 through the same potential difference will be
- A$\lambda \frac{m}{M}$
- ✓$\lambda \sqrt {\frac{m}{M}} $
- C$\lambda \frac{M}{m}$
- D$\lambda \sqrt {\frac{M}{m}} $
