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 min is the smallest possible wavelength of X-ray in the spectrum, the variation of log \lambda min with log V is correctly represented in:

Dual Nature of Radiation PYQ — JEE Main 2017 · Shift 0 · Q55 with Solution

View written solutionFree

Correct answer: B

When electrons are accelerated through a potential difference $V$ , their maximum kinetic energy is

$K_{\max}=eV$

If this entire energy is converted into a single X-ray photon, then

$h\nu_{\max}=eV$

Using $\nu = \dfrac{c}{\lambda}$ ,

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

Hence,

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

$\log \lambda_{\min} = \log\left(\frac{hc}{e}\right) - \log V$

This is of the form

$y = c - x$

where

So the graph of $\log \lambda_{\min}$ versus $\log V$ is a straight line with negative slope equal to $-1$ .

Therefore, the correct option is the one showing a straight line decreasing linearly.

So the correct option is B.