MCQ
An electron accelerated through a potential difference $V_1$ has a de-Broglie wavelength of $\lambda$. When the potential is changed to $V _2$, its de-Broglie wavelength increases by $50 \%$. The value of $\left(\frac{ V _1}{ V _2}\right)$ is equal to :
- A$3$
- ✓$\frac{9}{4}$
- C$\frac{3}{2}$
- D$4$




