Question

An electron beam is accelerated by a potential difference $V$ to hit a metallic target to produce $X-$rays. It produces continuous as well as characteristic $X$-rays. If $\lambda_{\text{min}}$ is the smallest possible wavelength of $X$-ray in the spectrum, the variation of $\log \, \lambda_{\text{min}}$ with log $V$ is correctly represented in :

• A.

  

• B.

  

• C.

  

• D.

  

Answer 1

The Correct Option is A

The problem involves understanding the relationship between the smallest wavelength of X-rays produced by an electron beam and the accelerating potential. This explanation will go through the concept and calculations needed to find that relationship.

When electrons are accelerated to hit a metallic target, the energy of the electrons is given by the formula:

E = eV where e is the charge of the electron and V is the potential difference.

The energy is also related to the wavelength of the X-rays produced, particularly the minimum wavelength, by the equation:

E = \frac{hc}{\lambda_{\text{min}}} where h is Planck's constant, c is the speed of light, and \lambda_{\text{min}} is the minimum wavelength.

Equating both expressions for energy:

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

Rearranging for \lambda_{\text{min}}, we get:

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

Now, taking the logarithm on both sides gives:

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

Which simplifies to:

\log \lambda_{\text{min}} = \log(hc) - \log(e) - \log(V)

Historically, this establishes a linear relationship between \log \lambda_{\text{min}} and \log V with a slope of -1. Therefore, the graph of \log \lambda_{\text{min}} vs. \log V is a straight line with a negative slope.

Based on these calculations, option (a) is correct, which represents this linear relationship.