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

$$3$$

B

$$\frac{9}{4}$$

C

$$\frac{3}{2}$$

D

$$4$$