Question

The minimum wavelength of X-rays produced by an electron accelerated through a potential difference of V volts is proportional to:

• A. \(V^2\)
• B. \({\sqrt{V}}\)
• C. \(\frac{1}{V}\)
• D. \(\frac{1}{\sqrt{V}}\)

Answer 1

The Correct Option is C

To determine the relationship between the minimum wavelength of X-rays and the potential difference through which electrons are accelerated, we use the concept of X-ray production in a Coolidge tube. When electrons are accelerated through a potential difference \( V \), their kinetic energy becomes \( eV \), where \( e \) is the charge of the electron. Upon collision with a target, this energy can be converted into a photon of X-ray, obeying the energy conservation principle. The energy of a photon is given by \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength. Thus, equating the energy, we have:

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

Rearranging to find the minimum wavelength \(\lambda_{\text{min}}\), we obtain:

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

This expression shows that the minimum wavelength is inversely proportional to the potential difference \( V \). Therefore, the correct proportionality for the minimum wavelength \(\lambda_{\text{min}}\) is:

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