JEE MAIN - Physics (2017 (Offline) - No. 12)
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:
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Explanation
In X-ray tube, $${\lambda _{\min }} = {{hc} \over {eV}}$$
$$\log \left( {{\lambda _{\min }}} \right) = \log \left( {{{hc} \over e}} \right) - \log V$$
Clearly, log ($$\lambda $$min) versus log V graph slope is negative hence option (b) is correct.
$$\log \left( {{\lambda _{\min }}} \right) = \log \left( {{{hc} \over e}} \right) - \log V$$
Clearly, log ($$\lambda $$min) versus log V graph slope is negative hence option (b) is correct.
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