The energy that should be added to an electron to reduce its de Broglie wavelength from 1 nm to 0.5 nm is:

thrice the initial energy

de Broglie wavelength,

$λ=\frac{h}{p}=\frac{h}{\sqrt{2mE}}$ (i)

After decreasing wavelength,

$λ'=\frac{h}{p'}=\frac{h}{\sqrt{2mE'}}$ (ii)

From eqs. (i) and (ii),

$\frac{λ}{λ'}=\sqrt{\frac{E}{E'}}$

Putting values of λ' and λ, we get

$\frac{E}{E'}=(\frac{0.5}{1})^2$

$∴E'=\frac{E}{0.25}=4E$

The energy should be added to decrease the wavelength = $E'-E=4E-E=3E$