Question

Electrons of mass $m$ with de-Broglie wavelength $\lambda$ fall on the target in an X-ray tube. The cutoff wavelength $(\lambda_0)$ of the emitted X-ray is

• A. $\lambda_0 = \frac{2 mc \lambda^2}{h}$
• B. $\lambda_0 = \frac{2h}{mc}$
• C. $\lambda_0 = \frac{2 m^2c^2 \lambda^3}{h^2}$
• D. $\lambda_0 = \lambda$

Answer 1

The Correct Option is A

To determine the cutoff wavelength (\lambda_0) of the emitted X-ray in an X-ray tube, we need to understand the relationship between the kinetic energy of the incident electrons and the energy of the emitted X-rays.

The kinetic energy (KE_e) of the electrons can be expressed in terms of their wavelength using the de-Broglie relation:

KE_e = \frac{h^2}{2m\lambda^2}

where h is Planck's constant, m is the mass of the electron, and \lambda is the de-Broglie wavelength.

In an X-ray tube, the maximum energy of the emitted X-ray photon is equal to the kinetic energy of the incident electron, which corresponds to the cutoff wavelength (\lambda_0). This is given by:

E_{photon} = \frac{hc}{\lambda_0}

where c is the speed of light.

Setting the maximum photon energy equal to the electron's kinetic energy, we have:

\frac{hc}{\lambda_0} = \frac{h^2}{2m\lambda^2}

Solving for \lambda_0, we rearrange the equation:

\lambda_0 = \frac{2mc\lambda^2}{h}

Thus, the cutoff wavelength (\lambda_0) of the emitted X-ray is \frac{2mc\lambda^2}{h}. This corresponds to option 1.

Let's briefly analyze why the other options are incorrect:

• \lambda_0 = \frac{2h}{mc}: This expression doesn't incorporate the de-Broglie wavelength \lambda that relates to the electron, thus it's not applicable here.
• \lambda_0 = \frac{2 m^2c^2 \lambda^3}{h^2}: Incorrect power relations between mass, speed of light, and Planck's constant are being used.
• \lambda_0 = \lambda: Assumes that the cutoff and de-Broglie wavelengths are the same, which is not correct as the cutoff wavelength depends on the energy transformation happening in the X-ray tube.