An X-ray tube used in medical diagnostic imaging produces continuous X-rays with a minimum… — JAMB UTME

An X-ray tube used in medical diagnostic imaging produces continuous X-rays with a minimum cutoff wavelength of $0.04125\text{ nm}$ . Given that Planck's constant $h = 6.6 \times 10^{-34}\text{ J}\cdot\text{s}$ , the speed of light $c = 3.0 \times 10^{8}\text{ m/s}$ , and the elementary charge $e = 1.6 \times 10^{-19}\text{ C}$ , what is the accelerating potential difference across the tube in kilovolts (kV)?

Answer: 30 kV

Answer

The accelerating potential difference across the X-ray tube is 30 kV.

By Duane-Hunt's law, the maximum energy of an emitted X-ray photon corresponds to the complete conversion of an electron's kinetic energy acquired across potential difference

V

e V = \frac{h c}{\lambda_{\text{min}}}

. Substituting

h = 6.6 \times 10^{-34}\text{ J}\cdot\text{s}

c = 3.0 \times 10^8\text{ m/s}

e = 1.6 \times 10^{-19}\text{ C}

, and

\lambda_{\text{min}} = 4.125 \times 10^{-11}\text{ m}

yields

V = 30,000\text{ V} = 30\text{ kV}

Step-by-Step Solution

Convert the given wavelength into SI units of meters

\lambda_{\text{min}} = 4.125 \times 10^{-11}\text{ m}

Standard SI units are required for calculations involving physical constants.

State the Duane-Hunt relation for maximum photon energy

e V = E_{\text{max}} = \frac{h c}{\lambda_{\text{min}}}

The maximum photon energy occurs when all kinetic energy of an accelerating electron is converted into a single X-ray photon.

Rearrange the equation and substitute the numerical values to solve for

V

V = \frac{(6.6 \times 10^{-34})(3.0 \times 10^{8})}{(1.6 \times 10^{-19})(4.125 \times 10^{-11})} = 30,000\text{ V}

Evaluates the required accelerating potential difference in volts.

Express the final voltage in kilovolts (kV)

V = 30\text{ kV}

The question asks for the answer specifically in kilovolts.

Key Concept

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

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