X Rays and the Ordering of the Elements — Question 1

Section 40-9 · Physics pages 6732–6733

Question 1 The X-ray continuum cutoffdepends on accelerating voltage Electrons initially at rest are accelerated through a potential difference V and strike a massive target. A single photon can take at most the incident electron kinetic energy; neglect target recoil. Use hc = 1240 eV nm. λ I λmin Schematic; labels, not drawing lengths, specify the data. (a) Derive the minimum emitted wavelength and evaluate it at V = 40.0 kV. (b) Find the voltage required to reduce that minimum wavelength by 25 percent.
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Question 1 - Solution The X-ray continuum cutoffdepends on accelerating voltage λ I λmin (a) Method and calculation K = eV, Eγ,max = eV ⇒λmin = hc/(eV ) . λmin = 1240/40000 = .0310 nm . (b) Interpretation and further calculation λmin′ = .75λmin, V ′/V = λmin/λmin′. V ′ = 40.0/.75 = 53.33 kV . Physical check The cutoffdescribes the bremsstrahlung continuum and follows total energy conservation. Characteristic line wavelengths depend primarily on atomic structure and need not shift when the voltage changes.

Question 1

The X-ray continuum cutoff depends on accelerating voltage

Electrons initially at rest are accelerated through a potential difference V and strike a massive target. A single photon can take at most the incident electron kinetic energy; neglect target recoil. Use hc=1240eVnmhc=1240\,\mathrm{eV\,nm}.

Diagram shown in the original SVG worksheet.

  1. Derive the minimum emitted wavelength and evaluate it at V=40.0kVV=40.0\,\mathrm{kV}.

  2. Find the voltage required to reduce that minimum wavelength by 25 percent.

Question 1 - Solution

The X-ray continuum cutoff depends on accelerating voltage

Diagram shown in the original SVG worksheet.

(a) Method and calculation

K=eV,Eγ,max=eV⇒λmin=hc/(eV).K=eV,\quad E_{\gamma,\max}=eV\Rightarrow\boxed{\lambda_{\min}=hc/(eV)}.λmin=1240/40000=.0310nm.\boxed{\lambda_{\min}=1240/40000=.0310\,\mathrm{nm}}.

(b) Interpretation and further calculation

λmin′=.75λmin,V′/V=λmin/λmin′.\lambda_{\min}\prime=.75\lambda_{\min},\quad V\prime/V=\lambda_{\min}/\lambda_{\min}\prime.V′=40.0/.75=53.33kV.\boxed{V\prime=40.0/.75=53.33\,\mathrm{kV}}.

Physical check

The cutoff describes the bremsstrahlung continuum and follows total energy conservation. Characteristic line wavelengths depend primarily on atomic structure and need not shift when the voltage changes.