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Difficulty: MediumX-rays: Production, Properties, and Applications

An X-ray tube is operated at an accelerating potential difference of 15.0 kV15.0\text{ kV}. Assuming all the kinetic energy of an electron is transferred into a single photon upon collision with the target, what is the minimum cutoff wavelength (\(\lambda_{\min}\)) of the emitted X-rays?

(Take Planck's constant h=6.60×10−34 J sh = 6.60 \times 10^{-34}\text{ J s}, speed of light c=3.00×108 m s−1c = 3.00 \times 10^8\text{ m s}^{-1}, and elementary charge e=1.60×10−19 Ce = 1.60 \times 10^{-19}\text{ C})

  1. 8.25×10−11 m8.25 \times 10^{-11}\text{ m}Answer
  2. B
    1.32×10−14 m1.32 \times 10^{-14}\text{ m}
  3. C
    1.21×1010 m−11.21 \times 10^{10}\text{ m}^{-1}
  4. D
    8.25×10−8 m8.25 \times 10^{-8}\text{ m}

Answer

The minimum cutoff wavelength of the emitted X-rays is 8.25×10−11 m8.25 \times 10^{-11}\text{ m}.
The correct answer is derived using Duane-Hunt's equation λmin⁡=hceV\lambda_{\min} = \frac{hc}{eV}. Substituting the known values (h=6.60×10−34 J sh = 6.60 \times 10^{-34}\text{ J s}, c=3.00×108 m s−1c = 3.00 \times 10^8\text{ m s}^{-1}, e=1.60×10−19 Ce = 1.60 \times 10^{-19}\text{ C}, V=15.0×103 VV = 15.0 \times 10^3\text{ V}) gives λmin⁡=8.25×10−11 m\lambda_{\min} = 8.25 \times 10^{-11}\text{ m}.

Step-by-Step Solution

1
Identify given parameters and convert to SI units.
Accelerating potential V=15.0 kV=15.0×103 VV = 15.0\text{ kV} = 15.0 \times 10^3\text{ V}, Planck's constant h=6.60×10−34 J sh = 6.60 \times 10^{-34}\text{ J s}, speed of light c=3.00×108 m s−1c = 3.00 \times 10^8\text{ m s}^{-1}, charge of electron e=1.60×10−19 Ce = 1.60 \times 10^{-19}\text{ C}.
SI units are mandatory to ensure dimensional correctness during photon energy evaluation.
2
Apply the Duane-Hunt Law for the maximum photon energy and minimum wavelength.
Emax⁡=eV=hcλmin⁡  ⟹  λmin⁡=hceVE_{\max} = e V = \frac{h c}{\lambda_{\min}} \implies \lambda_{\min} = \frac{h c}{e V}.
The shortest wavelength corresponds to maximum energy transfer when an electron yields all its kinetic energy to a single X-ray photon.
3
Substitute numerical values into the equation and calculate λmin⁡\lambda_{\min}.
\(\lambda_{\min} = \frac{6.60 \times 10^{-34} \times 3.00 \times 10^8}{1.60 \times 10^{-19} \times 15.0 \times 10^3} = \frac{1.98 \times 10^{-25}}{2.40 \times 10^{-15}} = 8.25 \times 10^{-11}\text{ m}\).
Carrying out the arithmetic yields the exact Duane-Hunt minimum cutoff wavelength.

Key Concept

Duane-Hunt Law and Cutoff Wavelength in X-ray Production
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Topics

X-rays: Production, Properties, and Applications

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