ANS= 0.018nm Please show ALL steps. Characteristic x rays: A beam of high energy

Question

ANS= 0.018nm

Please show ALL steps.

Characteristic x rays: A beam of high energy electrons is incident on a tantalum (atomic number 73) target One the electrons in the beam ejects K-shell (n = 1) electron from a tantalum atom An N-shell (n = 4) electron then release an x-ray photon as it makes a downward transition to fill the "hole" Estimate the wavelength of the X-ray photon produced. Remember The number of electrons in a closed shell is 2n^2 The energies of the various in multi-electron atoms increase In roughly the following order(don't worry about anomalies) 1s

Explanation / Answer

4)

We know that 1/ = RZ2 [ 1/n1^2 - 1/n2^2] , = wavelength of X-ray photon

atomic number of Tantalum Z = 73

Given that n1 = 1 , n2 = 4

Then,

1/ = Rz2 [1/n1^2 - 1/n2^2]

     = R(73)2 [ 1/1^2 - 1/4^2]

     = R (5329) [ 1- 1/16]

      = R (5329)(15/16)

1/ = R (5329) (15/16)

= (16/15) (5329) (1/R)

   = (16/15) (5329) ( 91.17 nm)      ( since R = 1.09678 x 10-2 nm-1 then 1/R = 91.17 nm)

   = 0.018 nm

wavelength = 0.018 nm

Therefore,

wavelength of X -ray photon = 0.018 nm

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