Q 12-11-174JEE MainJEE Main 2017 (2 Apr)Easy
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:
Answer: (B) see figure
The shortest wavelength corresponds to an electron giving all its energy $eV$ to one photon:
$$\frac{hc}{\lambda_{\min}} = eV \;\Rightarrow\; \lambda_{\min} = \frac{hc}{eV}$$
$$\log\lambda_{\min} = \log\frac{hc}{e} - \log V$$
This is a straight line of slope $-1$ with a positive intercept: a falling straight line, graph (2).
Solution by Sreeraj P, M.Sc Physics