An X-ray tube operates at an initial potential difference of V_1. When the potential diffe… — JAMB UTME

An X-ray tube operates at an initial potential difference of $V_1$ . When the potential difference is increased by $20.0\text{ kV}$ , the minimum cutoff wavelength of the emitted X-rays is reduced to one-third of its initial value. What is the initial operating potential difference $V_1$ of the X-ray tube?

$10.0\text{ kV}$ Answer
B
$30.0\text{ kV}$
C
$40.0\text{ kV}$
D
$5.0\text{ kV}$

Answer

The initial operating potential difference of the tube is

10.0\text{ kV}

The Duane-Hunt law states that

\lambda_{\text{min}} = \frac{hc}{eV}

. Because minimum wavelength is inversely proportional to potential difference, reducing the minimum wavelength to one-third requires the operating potential to increase by a factor of three (

V_2 = 3V_1

). Given that

V_2 = V_1 + 20.0\text{ kV}

, setting

V_1 + 20.0\text{ kV} = 3V_1

gives

2V_1 = 20.0\text{ kV}

, which solves to

V_1 = 10.0\text{ kV}

Step-by-Step Solution

Relate the minimum cutoff wavelength to potential difference using the Duane-Hunt law.

\lambda_{\text{min}} = \frac{hc}{eV}

, which implies

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

The maximum photon energy equals the kinetic energy of the incident electrons.

Set up the ratio between initial and final conditions.

Since

\lambda_2 = \frac{1}{3}\lambda_1

, the new voltage must be three times the initial voltage:

V_2 = 3V_1

Cutoff wavelength is inversely proportional to accelerating voltage.

Substitute

V_2 = V_1 + 20.0\text{ kV}

into the equation

V_2 = 3V_1

and solve for

V_1

V_1 + 20.0\text{ kV} = 3V_1 \implies 2V_1 = 20.0\text{ kV} \implies V_1 = 10.0\text{ kV}

Solving the linear algebraic equation yields the initial potential difference.

Key Concept

Duane-Hunt Law and Inverse Relationship between Minimum Wavelength and Accelerating Potential

Estimated Time:2m 0s

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