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

The correct answer is Option (4) → thrice the initial energy

According to De-Broglie wavelength,

$λ=\frac{h}{p}=\frac{h}{\sqrt{2mE}}$

After decreasing wavelength,

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

$∴\frac{λ}{λ'}=\sqrt{\frac{E'}{E}}$

$⇒\frac{E'}{E}=\left(\frac{λ}{λ'}\right)^2$

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

$⇒\frac{E'}{E}=4$

$⇒E'=4E$

∴ energy which should be added is,

$E=E'-E$

$=4E-E$

$=3E$