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Zorluk: ZorX-rays: Production, Properties, and Applications

In a Coolidge X-ray tube, electrons are accelerated from rest through an unknown potential difference VV towards a target metal. If the shortest cutoff wavelength of the resulting continuous X-ray spectrum is recorded as 0.0414 nm0.0414\text{ nm}, what is the operating potential difference VV of the tube?

(Take Planck's constant h=6.63×10−34 J sh = 6.63 \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}, elementary charge e=1.60×10−19 Ce = 1.60 \times 10^{-19}\text{ C})

  1. 30.0 kV30.0\text{ kV}Cevap
  2. B
    15.0 kV15.0\text{ kV}
  3. C
    300 V300\text{ V}
  4. D
    60.0 kV60.0\text{ kV}

Cevap

The operating potential difference VV of the tube is 30.0 kV30.0\text{ kV}.
According to the Duane-Hunt law for X-ray emission, the shortest wavelength λmin⁡\lambda_{\min} corresponds to the maximum kinetic energy acquired by an electron accelerated through a voltage VV. The mathematical relationship is eV=hcλmin⁡e V = \frac{h c}{\lambda_{\min}}. Substituting h=6.63×10−34 J sh = 6.63 \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}, and λmin⁡=4.14×10−11 m\lambda_{\min} = 4.14 \times 10^{-11}\text{ m} gives V=1.989×10−256.624×10−30=30.0 kVV = \frac{1.989 \times 10^{-25}}{6.624 \times 10^{-30}} = 30.0\text{ kV}.

Adım Adım Çözüm

1
Convert the given minimum wavelength from nanometers to meters
λmin⁡=0.0414 nm=0.0414×10−9 m=4.14×10−11 m\lambda_{\min} = 0.0414\text{ nm} = 0.0414 \times 10^{-9}\text{ m} = 4.14 \times 10^{-11}\text{ m}
SI unit consistency requires length to be in meters.
2
Apply the Duane-Hunt law for continuous X-ray production
Emax⁡=eV=hfmax⁡=hcλmin⁡E_{\max} = e V = h f_{\max} = \frac{h c}{\lambda_{\min}}
The maximum photon energy equals the kinetic energy of the incident electron.
3
Rearrange the equation to solve for the accelerating voltage VV
V=hceλmin⁡V = \frac{h c}{e \lambda_{\min}}
Isolating the operating potential difference VV on one side of the equation.
4
Substitute the physical constants and calculate VV
V=(6.63×10−34 J s)(3.00×108 m s−1)(1.60×10−19 C)(4.14×10−11 m)=1.989×10−256.624×10−30=30,012 V≈30.0 kVV = \frac{(6.63 \times 10^{-34}\text{ J s}) (3.00 \times 10^8\text{ m s}^{-1})}{(1.60 \times 10^{-19}\text{ C}) (4.14 \times 10^{-11}\text{ m})} = \frac{1.989 \times 10^{-25}}{6.624 \times 10^{-30}} = 30,012\text{ V} \approx 30.0\text{ kV}
Performing numeric evaluation gives the operating potential difference in kilovolts.

Anahtar Kavram

Duane-Hunt Law and Cutoff Wavelength in X-ray Tubes
Tahmini Süre:2m 0s
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Konular

X-rays: Production, Properties, and Applications

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