The correct option is C becomes half
de-Broglie wavelength is given by λ=hp
where, momentum of the particle p=√2mK
Kinetic energy of the particle K=eV
⟹ λ=h√2meV
We get λ∝1√V
Thus de-Broglie wavelength of the electron will become half if the accelerating potential is increased by a factor of 4.