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

Sure, to calculate the de-Broglie wavelength of an electron, we can use the formula:

λ = h / p

where:

• λ is the de-Broglie wavelength,
• h is the Planck constant (6.62607015 × 10^-34 m^2 kg / s),
• p is the momentum of the particle.

The momentum of a particle can be calculated from its kinetic energy (E) using the formula:

p = sqrt(2mE)

where:

• m is the mass of the electron (9.10938356 × 10^-31 kilograms),
• E is the kinetic energy.

Given that the kinetic energy E is 1200 keV, we first need to convert this energy to Joules (J) because the other constants are in SI units. 1 eV = 1.60218 x 10^-19 J, so:

E = 1200 x 10^3 eV = 1200 x 10^3 x 1.60218 x 10^-19 J = 1.922616 x 10^-13 J

Now we can calculate the momentum:

p = sqrt(2 * 9.10938356 × 10^-31 kg * 1.922616 x 10^-13 J) = 1.64424 x 10^-23 kg m/s

Finally, we can calculate the de-Broglie wavelength:

λ = h / p = 6.62607015 × 10^-34 m^2 kg / s / 1.64424 x 10^-23 kg m/s = 4.03 x 10^-11 m

So, the de-Broglie wavelength of an electron with 1200 keV kinetic energy is approximately 4.03 x 10^-11 meters.