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

Step-by-step solution

The de-Broglie wavelength of an electron is given by = h 2mK Squaring both sides, we get K = h^2 2m ^2 This implies K 1 ^2 For the initial state, ₁ = 1 nm, so K₁ 1 1^2 = 1 For the final state, ₂ = 0.5 nm, so K₂ 1 (0.5)^2 = 4 Therefore, K₂ = 4K₁ The additional energy required is K = K₂ - K₁ = 4K₁ - K₁ = 3K₁ Given that the additional energy is x times the initial kinetic energy, we have x = 3 Answer: 3