An X-ray tube operates at an accelerating potential difference of 25.0\text{ kV}. Calculat… — JAMB UTME

An X-ray tube operates at an accelerating potential difference of $25.0\text{ kV}$ . Calculate the maximum frequency of the emitted X-ray radiation in units of $10^{18}\text{ Hz}$ . (Take Planck's constant $h = 6.63 \times 10^{-34}\text{ J}\cdot\text{s}$ and elementary charge $e = 1.60 \times 10^{-19}\text{ C}$ ).

Answer: 6.03 10^18 Hz

Answer

The maximum frequency of the emitted X-rays is

6.03 \times 10^{18}\text{ Hz}

, which gives a value of

6.03

in units of

10^{18}\text{ Hz}

The maximum frequency of X-ray photons emitted occurs when an accelerating electron transfers all of its kinetic energy (

e V

) into a single photon (

h f_{\text{max}}

). Substituting

e = 1.60 \times 10^{-19}\text{ C}

V = 25,000\text{ V}

, and

h = 6.63 \times 10^{-34}\text{ J}\cdot\text{s}

gives

f_{\text{max}} = 6.03 \times 10^{18}\text{ Hz}

, which equals

6.03

in the requested unit of

10^{18}\text{ Hz}

Step-by-Step Solution

Convert potential difference from kilovolts to volts

V = 25.0\text{ kV} = 25,000\text{ V}

Standard SI units are required for calculations.

Determine maximum kinetic energy of the incident electrons

E_{\text{max}} = e V = 1.60 \times 10^{-19}\text{ C} \times 25,000\text{ V} = 4.00 \times 10^{-15}\text{ J}

The maximum photon energy produced equals the full kinetic energy acquired by an accelerated electron.

Calculate maximum frequency

f_{\text{max}}

using Duane-Hunt relation

f_{\text{max}} = \frac{E_{\text{max}}}{h} = \frac{4.00 \times 10^{-15}\text{ J}}{6.63 \times 10^{-34}\text{ J}\cdot\text{s}} \approx 6.03 \times 10^{18}\text{ Hz}

According to the Duane-Hunt law,

e V = h f_{\text{max}}

Key Concept

Duane-Hunt Law and Maximum X-ray Frequency

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