An X-ray tube produces a continuous spectrum of radiation with its short wavelength end0.65Å. Determine the maximum energy of a photon in the radiation.a.19.13 keV.b.1.913 keV.c.18.93 keVd.19.38 keV.

The energy of a photon can be calculated using the formula:

E = hc/λ

where: E is the energy of the photon, h is Planck's constant (6.626 x 10^-34 Joule second), c is the speed of light (3 x 10^8 m/s), and λ is the wavelength of the photon.

Given that the shortest wavelength end of the spectrum is 0.65Å (Angstroms), we first need to convert this to meters because the speed of light is given in meters per second.

1Å = 1 x 10^-10 m, so 0.65Å = 0.65 x 10^-10 m = 6.5 x 10^-11 m.

Substituting these values into the formula gives:

E = (6.626 x 10^-34 J.s)(3 x 10^8 m/s) / 6.5 x 10^-11 m E = 3.05 x 10^-15 Joule

To convert this energy to electron volts (eV), we use the conversion factor 1eV = 1.6 x 10^-19 Joule:

E = 3.05 x 10^-15 J * (1 eV / 1.6 x 10^-19 J) E = 19.06 x 10^3 eV E = 19.06 keV

So, the maximum energy of a photon in the radiation is approximately 19.06 keV. The closest answer to this is 19.13 keV, so the answer is (a) 19.13 keV.