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
    $4$

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