The additional energy that should be given to an electron to reduce its de-Broglie wavelength from 1 nm to 0.5 nm is $x$ times the initial kinetic energy. The value of $x$ is

The de-Broglie wavelength of an electron is given by $\lambda = \dfrac{h}{\sqrt{2mK}}$

Squaring both sides, we get $K = \dfrac{h^2}{2m\lambda^2}$

This implies $K \propto \dfrac{1}{\lambda^2}$

For the initial state, $\lambda_1 = 1$ nm, so $K_1 \propto \dfrac{1}{1^2} = 1$

For the final state, $\lambda_2 = 0.5$ nm, so $K_2 \propto \dfrac{1}{(0.5)^2} = 4$

Therefore, $K_2 = 4K_1$

The additional energy required is $\Delta K = K_2 - K_1 = 4K_1 - K_1 = 3K_1$

Given that the additional energy is $x$ times the initial kinetic energy, we have $x = 3$

Answer: $3$