What is the de-Broglie wave-length of an electron with 1200keV kinetic energy

Sure, let's solve this step by step.

Step 1: Convert the kinetic energy from keV to Joules. We know that 1 eV = 1.6 x 10^-19 Joules. Therefore, 1200 keV = 1200 x 10^3 x 1.6 x 10^-19 Joules = 1.92 x 10^-14 Joules.

Step 2: Use the de-Broglie wavelength formula. The de-Broglie wavelength formula is λ = h / p, where h is Planck's constant and p is the momentum of the particle.

Step 3: Calculate the momentum. We know that the kinetic energy (E) = p^2 / 2m, where m is the mass of the electron. Therefore, p = sqrt(2mE). The mass of an electron (m) is 9.1 x 10^-31 kg. So, p = sqrt(2 * 9.1 x 10^-31 kg * 1.92 x 10^-14 Joules) = 4.37 x 10^-23 kg m/s.

Step 4: Substitute the values into the de-Broglie wavelength formula. Planck's constant (h) is 6.63 x 10^-34 Js. Therefore, λ = h / p = 6.63 x 10^-34 Js / 4.37 x 10^-23 kg m/s = 1.52 x 10^-11 m.

So, the de-Broglie wavelength of an electron with 1200 keV kinetic energy is approximately 1.52 x 10^-11 m.