Correct Answer  Option 2 : (1/√2) times
CONCEPT:

de Broglie wavelength of electrons: Louis de Broglie theorized that not only light possesses both wave and particle properties, but rather particles with mass  such as electrons  do as well.
 The wavelength of material waves is also known as the de Broglie wavelength.

de Broglie wavelength (λ) of electrons can be calculated from Planks constant h divided by the momentum of the particle.
λ = h/p
where λ is de Broglie wavelength, h is Plank's const, and p is the momentum.
 In terms of Energy, de Broglie wavelength of electrons:
\(λ=\frac{h}{\sqrt{2mE}}\)
where λ is de Broglie wavelength, h is Plank's const, m is the mass of an electron, and E is the energy of the electron.
CALCULATION:
Given that E2 = 2E1
\(λ_1=\frac{h}{\sqrt{2mE_1}}\) .....(i)
\(λ_2=\frac{h}{\sqrt{2mE_2}}\) .....(ii)
\(\frac{λ_1}{λ_2}=\sqrt\frac{E_2}{E_1}=\sqrt\frac{2E_1}{E_1}\)
\(\frac{λ_2}{λ_1}=\sqrt\frac{1}{2}\)
\({λ_2}={λ_1}\frac{1}{\sqrt2}\)
So the correct answer is option 2.