The kinetic energy of a free electron increases to $3$ times the previous kinetic energy $(K.E.)$. The ratio of the new de-Broglie wavelength to the previous de-Broglie wavelength is:

  • A
    $\frac{1}{\sqrt{3}}$
  • B
    $\frac{1}{3}$
  • C
    $3$
  • D
    $\sqrt{3}$

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