X-Rays – JEE Main Previous Year Questions with Solutions

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Q. If a source of power 4kW produces $10^{20}$ photons/second, the radiation belongs to apart of the spectrum called :- (1) $\gamma$-rays              (2) X-rays              (3) ultraviolet rays              (4) microwaves [AIEEE – 2010]

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Sol. (2) No. of photons emitting per second from a source of power $\mathrm{P}$ is $\mathrm{n}=\left(5 \times 10^{24}\right) \mathrm{P} \lambda$ $\Rightarrow$ wavelength emitting $\lambda=\frac{\mathrm{n}}{\left(5 \times 10^{24}\right) \mathrm{P}}\left[\text { or } \lambda=\frac{\mathrm{nhc}}{\mathrm{P}}\right]$ And this wavlength comes in X ray region.

Q. 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 : [JEE Main-2017]

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Sol. (3) $\frac{\mathrm{hc}}{\lambda_{\min }}=\mathrm{eV}$ $\frac{1}{\lambda_{\text {min }}}=\frac{\mathrm{eV}}{\mathrm{hc}}$ $\ln \left(\frac{1}{\lambda_{\min }}\right)=\ell \mathrm{nV}+\ell \mathrm{n} \frac{\mathrm{e}}{\mathrm{hc}}$ $-\ell n\left(\lambda_{\min }\right)=\ell n V+\ell n \frac{\mathrm{e}}{\mathrm{hc}}$ $\ln \left(\lambda_{\min }\right)=-\ell \mathrm{n} \mathrm{V}-\ell \mathrm{n}\left(\frac{\mathrm{e}}{\mathrm{hc}}\right)$ It is a straight line with –ve slope.


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