An X-ray tube operates at an accelerating potential of 50.0\text{ kV}. Given Planck's cons… — JAMB UTME

An X-ray tube operates at an accelerating potential of $50.0\text{ kV}$ . Given Planck's constant $h = 6.63 \times 10^{-34}\text{ J s}$ , the speed of light $c = 3.00 \times 10^8\text{ m s}^{-1}$ , and the elementary charge $e = 1.60 \times 10^{-19}\text{ C}$ , what is the minimum wavelength $\lambda_{\min}$ of the emitted continuous X-rays, and how does this cutoff wavelength respond to an increase in the anode potential?

$2.49 \times 10^{-11}\text{ m}$ , and $\lambda_{\min}$ decreases when the anode potential is increased.Answer
B
$2.49 \times 10^{-8}\text{ m}$ , and $\lambda_{\min}$ decreases when the anode potential is increased.
C
$2.49 \times 10^{-11}\text{ m}$ , and $\lambda_{\min}$ remains unchanged because $\lambda_{\min}$ depends only on the filament current.
D
$4.02 \times 10^{-11}\text{ m}$ , and $\lambda_{\min}$ increases when the anode potential is increased.

Answer

The minimum wavelength is

2.49 \times 10^{-11}\text{ m}

, and

\lambda_{\min}

decreases when the anode potential is increased.

By the Duane-Hunt law, the maximum photon energy produced by electron impact equals the kinetic energy of the incident electron:

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

. Substituting

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

c = 3.00 \times 10^8\text{ m s}^{-1}

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

, and

V = 5.00 \times 10^4\text{ V}

yields

\lambda_{\min} = 2.49 \times 10^{-11}\text{ m}

. Because

\lambda_{\min}

is inversely proportional to

V

, increasing the anode potential decreases the minimum wavelength, making the beam more penetrating (harder).

Step-by-Step Solution

Convert the operating voltage to standard SI units (volts).

V = 50.0\text{ kV} = 50.0 \times 10^3\text{ V} = 5.00 \times 10^4\text{ V}

Equations involving fundamental constants require input quantities in SI base units.

Apply the Duane-Hunt law for maximum photon energy / minimum wavelength in continuous X-ray production.

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

The maximum kinetic energy of an accelerating electron is completely converted into a single photon of minimum wavelength.

Substitute the physical values into the minimum wavelength formula.

$\lambda_{\min} = \frac{(6.63 \times 10^{-34}\text{ J s})(3.00 \times 10^8\text{ m s}^{-1})}{(1.60 \times 10^{-19}\text{ C})(5.00 \times 10^4\text{ V})} = \frac{1.989 \times 10^{-25}}{8.00 \times 10^{-15}} = 2.48625 \times 10^{-11}\text{ m} \approx 2.49 \times 10^{-11}\text{ m}$.

Direct numerical evaluation yields the minimum cutoff wavelength.

Analyze the functional relationship between anode potential

V

and cutoff wavelength

\lambda_{\min}

Since

\lambda_{\min} \propto \frac{1}{V}

, an increase in

V

results in a decrease in

\lambda_{\min}

Higher potential imparts greater kinetic energy to striking electrons, enabling emission of higher-frequency (shorter-wavelength) photons.

Key Concept

Duane-Hunt Law and Control Parameters of X-ray Production

Estimated Time:2m 0s

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