Step 1: Using the de-Broglie wavelength formula.
The de-Broglie wavelength \( \lambda \) is related to the momentum \( p \) of the electron by the equation:
\[
\lambda = \frac{h}{p}
\]
Where \( h \) is Planck's constant. The energy of the electron is related to its momentum by:
\[
E = \frac{p^2}{2m}
\]
Step 2: Using the change in wavelength.
Since the wavelength reduces by half, the momentum must increase by a factor of \( \sqrt{2} \). Therefore, the energy increases by a factor of 3.
Thus, the correct answer is (C) \( 3E \).