In an X-ray tube, non-relativistic electrons accelerated from rest hit a target anode, pro… — JAMB UTME

In an X-ray tube, non-relativistic electrons accelerated from rest hit a target anode, producing continuous X-radiation with a minimum cut-off wavelength of $\lambda_0$ . If the operating potential difference across the tube is adjusted such that the maximum momentum of the colliding electrons increases by $50\%$ , what is the new cut-off wavelength of the emitted X-rays in terms of $\lambda_0$ ?

A
$\frac{2}{3}\lambda_0$
$\frac{4}{9}\lambda_0$ Answer
C
$\frac{9}{4}\lambda_0$
D
$\frac{3}{2}\lambda_0$

Answer

The new cut-off wavelength of the emitted X-rays is

\frac{4}{9}\lambda_0

According to the Duane-Hunt law, the maximum photon energy produced in continuous X-radiation equals the maximum kinetic energy of the striking electrons:

E_{\text{max}} = \frac{hc}{\lambda_{\min}} = E_k

. Expressing kinetic energy in terms of momentum yields

E_k = \frac{p^2}{2m}

, which gives

\lambda_{\min} = \frac{2mhc}{p^2}

. Therefore,

\lambda_{\min}

is inversely proportional to

p^2

. When momentum increases by

50\%

(

p_2 = 1.5 p_1 = \frac{3}{2}p_1

p^2

increases by a factor of

\frac{9}{4}

. Consequently, the new cut-off wavelength becomes

\frac{4}{9}\lambda_0

Step-by-Step Solution

Relate electron momentum to electron kinetic energy

The kinetic energy

E_k

of non-relativistic electrons of mass

m

with momentum

p

E_k = \frac{p^2}{2m}

Electrons are accelerated through potential difference

V

, gaining kinetic energy

E_k = e V = \frac{p^2}{2m}

Apply the Duane-Hunt law for minimum X-ray wavelength

\lambda_0 = \frac{hc}{E_k} = \frac{2mhc}{p_1^2}

The maximum energy of an X-ray photon corresponds to the shortest (cut-off) wavelength

\lambda_{\min} = \frac{hc}{E_k}

Calculate the updated momentum and new kinetic energy ratio

New momentum

p_2 = 1.5 p_1 = \frac{3}{2} p_1

, so

p_2^2 = \frac{9}{4} p_1^2

An increase of

50\%

means multiplying the initial momentum by

1 + 0.5 = 1.5 = \frac{3}{2}

Determine the new cut-off wavelength

\lambda_{\text{new}}

\lambda_{\text{new}} = \frac{2mhc}{p_2^2} = \frac{2mhc}{\frac{9}{4}p_1^2} = \frac{4}{9} \left(\frac{2mhc}{p_1^2}\right) = \frac{4}{9}\lambda_0

Since

\lambda_{\min}

is inversely proportional to

p^2

, scaling momentum by

\frac{3}{2}

reduces the cut-off wavelength by a factor of

\left(\frac{2}{3}\right)^2 = \frac{4}{9}

Key Concept

Duane-Hunt Law and Electron Kinetics

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic