What is the minimum wavelength (to four decimal places and in units of nm) of X-rays produced when an electron beam, whose electrons have kinetic energy 46 keV, is directed head-on into a lead target.

The energy of a photon is given by the equation E = hc/λ, where h is Planck's constant (4.135667696 x 10^-15 eV*s), c is the speed of light (3 x 10^8 m/s), and λ is the wavelength.

We can rearrange this equation to solve for λ: λ = hc/E.

Given that the kinetic energy of the electrons is 46 keV, we can substitute this value into the equation:

λ = (4.135667696 x 10^-15 eV*s * 3 x 10^8 m/s) / 46 x 10^3 eV

This gives us a wavelength in meters. To convert this to nanometers (nm), we multiply by 10^9.

λ = (4.135667696 x 10^-15 eV*s * 3 x 10^8 m/s * 10^9 nm/m) / 46 x 10^3 eV

After calculating, we get λ = 0.0271 nm. This is the minimum wavelength of X-rays produced when an electron beam with a kinetic energy of 46 keV is directed head-on into a lead target.