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

Step-by-step solution

Given, initial value of de-Broglie wavelength, ₁=1 ~nm =1 10⁻⁹ ~m Final value of de-Broglie wavelength, ₂=0.5 ~nm =0.5 10⁻⁹ ~m de-Broglie wavelength in terms of energy E is given as aligned & &= h 2 m E & & & 1 E & & ₁ ₂ &= E₂ E₁ aligned where, E₁ and E₂ are initial and final energy, 1 10⁻⁹ 0.5 10⁻⁹ = E₂ E₁ 2= E₂ E₁ E₂=4 E₁ Addition of energy =E₂-E₁=4 E₁-E₁=3 E₁ = thrice the initial energy