University:
Massachusetts Institute of Technology
Course:
3.091SC | Introduction to Solid State Chemistry
Academic year:

2024
Views:

58

Pages:

2
Author:

Oliver Schroeder

Session #17 - Homework Solutions Problem #1 You are operating an x-ray tube with a chromium (Cr) target by applying an acceleration potential (V) of 60 kV. Draw a schematic of the x-ray spectrum emitted by this tube; label on it three characteristic λs and give the numerical value of two of these. Solution A characteristic x-ray spectrum of Cr will show λSWL , Kβ , K α and the continuous spectrum or Bremsstrahlung. We may quantify λK and λSWL . α 24Cr: ν Kα = 3 R(Z - 1)2 4 = 3 4 × 1.097 × 107 (23)2 = 4.35 × 109 m-1 λK = 2.3 × 10-10 m α λSWL = hc 1.24 × 10-6 m = = 2.07 × 10−11 m 4 eV 6 × 10 Problem #2 (a) An X-ray tube with a silver (Ag) target at a plate voltage of 66 kV. Calculate the value of λSWL , the shortest wavelength. (b) Sketch the emission spectrum (intensity vs. wavelength) of the Ag target in part (a). On your sketch, indicate the relative positions of the K α , Kβ , L α , and Lβ lines and λSWL . It is not necessary to calculate the λ values of the Kβ , L α , and Lβ lines. (c) In one or two sentences explain the origin of the continuous spectrum. Solution (a) λSWL = hc 12400 = = 0.188 Å eV 66 × 103 (b) See sketch above in answer to problem #1. The L α and Lβ lines will appear to the right of the analogous K lines (at higher values of λ ), the L α to the right of the Lβ . (c) Incident electrons are deflected by the negative charge of electrons in the target. Any change in velocity (speed or direction or both) is an acceleration. Accelerating a charge emits radiation. The deflected electrons’ acceleration is NOT QUANTIZED. Thus, the spectrum is continuous. Problem #3 Determine the wavelength of λK for molybdenum (Mo). α Solution Mo: Z = 42; Kα → ni=2; nf =1; σ=1 ν ν ν ⎡1 1⎤ = R(Z - 1)2 ⎢ ⎥ 2 Kα ni2 ⎦⎥ ⎣⎢ nf Kα Kα λK α 1⎤ ⎡1 ⎡1⎤ = 1.097 × 107 ⎢ ⎥ (42 − 1)2 ⎢ ⎥ 2 ⎣m ⎦ ⎣1 22 ⎦ = 1.38 × 1010 m−1 = 1 νK = 7.25 × 10-11 m α Problem #4 Identify the element giving rise to K α with λ = 2.51 × 10-10 m . Solution 1 λK =ν α (Z − 1) = Kα ⎡1 1⎤ 3 = R(Z - 1)2 ⎢ ⎥ = R(Z - 1)2 × 2 2 4 ni ⎥⎦ ⎣⎢ nf 4 = 22 3× λ ×R Z = 23 (Vanadium)