X-Rays – JEE Main Previous Year Questions with Solutions

JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas.

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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]

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]

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.