What potential difference must be applied across an X-ray tube to generate X-rays with a wavelength of no less than 0.10 nm ? Additionally, what is the maximum energy of a photon of this X-ray in joules?

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Quantrex Academy · Accepted Answer

Correct answer: C. 12.4 kV , 2.0 × 10^ -15 J

The minimum wavelength of X-rays produced in an X-ray tube is given by

\lambda_\min = \dfrac{hc}{eV}

.

Rearranging for the potential difference

V

:

V = \dfrac{hc}{e\lambda_\min}

Using the value of

hc \approx 1240 \text{ eV}\cdot\text{nm}

:

V = \dfrac{1240 \text{ eV}\cdot\text{nm}}{e \times 0.10 \text{ nm}} = 12400 \text{ V} = 12.4 \text{ kV}

The maximum energy of an X-ray photon corresponds to the minimum wavelength:

E_\max = \dfrac{hc}{\lambda_\min}

Substituting the standard values in SI units:

E_\max = \dfrac{6.63 \times 10^{-34} \times 3 \times 10^8}{0.10 \times 10^{-9}}

E_\max = 19.89 \times 10^{-16} \text{ J} \approx 2.0 \times 10^{-15} \text{ J}

What potential difference must be applied across an X-ray tube to...

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