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 76% of its normal value. Now what is the minimum x-ray wavelength produced by the machine?
lmin = ___________nm

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

Emax =___________ eV

(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

Answer 1

Answer:

(a) 0.032 nm

(b) 39,235 eV

(c) 70,267.8 eV

Explanation:

(a) The energy of a photon can be calculated using:

E = hc/λ                  equation (1)

where:

h = 4.13*10^-15 eV.s

c = 3*10^8 m/s

λ = 0.024*10^-9 m

Thus:

E = (4.13*10^-15)*(3*10^8)/0.024*10^-9 = 51,625 eV

Then we calculate 76% of this estimated energy and determine the new wavelength:

$E_{new} = 0.76*51625 = 39,235 eV$

Using equation (1) to determine the new wavelength:

λ$_{new} = \frac{h*c}{E_{new} }$

λ$_{new}$ = (4.13*10^-15)*(3*10^8)/39235 = 3.15*10^-11 m = 0.032 nm

(b) As calculated in part (a), the maximum x-ray energy this machine can produce is $E_{new} = 0.76*51625 = 39,235 eV$

(c)  The energy of a Ka x-ray photon can be estimated using:

$E_{ka} = (10.2 eV)*(Z-1)^{2}$

where Z is the atomic number = 84.

$E_{ka} = (10.2 eV)*(84-1)^{2}$ = 70,267.8 eV