(a) A certain medical machine emits x-rays with a minimum wavelength of 0.024 nm

Question

(a) A certain medical machine emits x-rays with a minimum wavelength of 0.024 nm. One day, the machine has an electrical problem and the voltage applied to the x-ray tube decreases to 75% of its normal value. Now what is the minimum x-ray wavelength produced by the machine? min = nm Help Enter (b) What is the maximum x-ray energy this machine (with electrical problems) can produce? Emax = eV Help Enter (c) The atomic number of an element is 82. According to the Bohr model, what is the energy of a Ka x-ray photon? E = eV Help Enter

Explanation / Answer

here,

Minimumwavelength(?min) of X - ray =0.025nm = 0.025*10-9m

Nowfomrula for calculating the value of the max energy is Emax = (hc) / (Lmin)

where L is waveength

so Emax = = (6.625*10^-34)(3.0*10^8) /(0.024*10^-9) = 828.12 e -17 J

Also Now Energy K = Vq = hc/L

h is plancks Constant

c is speed of ight

L is waveelngth

so

part A :

where V is Potential Diff and q is charge

as K is Propoertiona to V

we get New K = 75% of E

Then E' =(0.75) * 828 e -17 = 621 e -17 J

New Wavelength L = hc/E = (6.625*10^-34* 3.0*10^8)/(621*10^-17)

L new = 0.032 nm

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b)

For photons of wavelength l,

E = hc/L

l =0.024e-9 m

E = 6.626e -34 * 3e 8/(0.024e -9)

E = 8.28 e -15/1.6 e -19

75 % of this = 51.75 * 0.75

E = 38.81 KeV

With the electrical problems, the maximum energy (which yields the shortest wavelength) goes to 38.81 KeV

c) The energy here = (Z-1)^2*(E2 - E1)

E = = 81^2*(-3.4eV --13.6eV)

E = 6.69*10^4

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