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

An X-ray tube operates at an accelerating potential difference of 20 kV20\text{ kV}. What is the maximum energy of the produced X-ray photons in Joules? (Take elementary charge e=1.6×10−19 Ce = 1.6 \times 10^{-19}\text{ C})

  1. 3.2×10−15 J3.2 \times 10^{-15}\text{ J}Answer
  2. B
    3.2×10−18 J3.2 \times 10^{-18}\text{ J}
  3. C
    1.25×1023 J1.25 \times 10^{23}\text{ J}
  4. D
    2.0×104 J2.0 \times 10^{4}\text{ J}

Answer

3.2×10−15 J3.2 \times 10^{-15}\text{ J}
According to the Duane-Hunt law, the maximum energy of an emitted X-ray photon equals the maximum kinetic energy gained by an electron accelerated through potential difference VV, which is Emax=eVE_{\text{max}} = e V. Substituting e=1.6×10−19 Ce = 1.6 \times 10^{-19}\text{ C} and V=20,000 VV = 20,000\text{ V} yields 3.2×10−15 J3.2 \times 10^{-15}\text{ J}.

Step-by-Step Solution

1
Convert the accelerating potential difference from kilovolts (kV) to volts (V).
V=20 kV=20,000 V=2.0×104 VV = 20\text{ kV} = 20,000\text{ V} = 2.0 \times 10^{4}\text{ V}
Standard SI units require potential difference in volts.
2
Apply the Duane-Hunt relationship for maximum photon energy Emax=eVE_{\text{max}} = e V.
Emax=1.6×10−19 C×2.0×104 V=3.2×10−15 JE_{\text{max}} = 1.6 \times 10^{-19}\text{ C} \times 2.0 \times 10^{4}\text{ V} = 3.2 \times 10^{-15}\text{ J}
The kinetic energy acquired by accelerated electrons is completely converted into the maximum energy of an emitted X-ray photon.

Key Concept

Duane-Hunt Law and Maximum X-ray Photon Energy
Estimated Time:45s
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Topics

X-rays: Production, Properties, and Applications

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