Question

Please Show work

1) A certain medical machine emits x-rays with a minimum wavelength of 0.025 nm. One day, the machine has an electrical problem and the voltage applied to the x-ray tube decreases to 74% of its normal value. Now what is the minimum x-ray wavelength produced by the machine?

?min = _________ nm

2) What is the maximum x-ray energy this machine (with electrical problems) can produce?

Emax = _________ eV

3) The atomic number of an element is 82. According to the Bohr model, what is the energy of a K? x-ray photon?

E = ___________ eV

Answer 1

Giventhat

Minimumwavelength(?_min) of X - ray =0.025nm

                                                          = 0.025*10^-9m

Nowfomrula for calculating the value of the max energy is

              E_max = (hc) / (?_min)

                      = (6.625*10^-34J.s)(3.0*10⁸m/s) /(0.025*10^-9m)

                      = 795*10^-17J

Sincewe know that formula energy (E) = qV

Here Vis the applied voltage.

Soenergy(E) is proportional to applied voltage(V).

Herethe applied voltage decreases to 74% of its normal value, then theenergy also decreases to 74% of its normal value.

Then E' =(0.74)E

             = (0.74)(795*10^-17J)

             = 588.3*10^-17J

Thennew wavelength is

          ?' = (hc) / E'

              = (6.625*10^-34J.s)(3.0*10⁸m/s) /(588.3*10^-17J) = 0.3378*10^-10 m = 0.3378 nm

b)

For photons of wavelength l,
E = hc/l

l =0.025e-9 m
h = 4.136e-15 eV*s

E = 4.136e-15 eV*s * 3e8 m/s / 25e-12 m
= 0.4966e5 eV = 49.66 KeV

With the electrical problems, the maximum energy (which yields the shortest wavelength) goes to 49.66 * .74 = 36.75 KeV.

c) The energy here = (Z-1)^2*(E2 - E1) = 81^2*(-3.4eV --13.6eV) = 6.69x10^4eV