Hard X-rays | Soft X-rays |
More penetration power | Less penetration power |
More frequency of the order of \[\approx {{10}^{19}}\,Hz\] | Less frequency of the order of \[\approx {{10}^{16}}\,Hz\] |
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(1) X-rays were discovered by scientist Rontgen thats why they are also called Rontgen rays.
(2) Rontgen discovered that when pressure inside a discharge tube is kept \[{{10}^{-3}}\,mm\] of Hg and potential difference is kept 25 kV, then some unknown radiations (X-rays) are emitted by anode.
(3) There are three essential requirements for the production of X-rays.
(i) A source of electron
(ii) An arrangement to accelerate the electrons
(iii) A target of suitable material of high atomic weight and high melting point on which these high speed electrons strike.
(1) The scattering of a photon by an electron is called Compton effect.
(2) The energy and momentum is conserved.
(3) Scattered photon will have less energy (more wavelength) as compare to incident photon (less wavelength).
(4) The energy lost by the photon is taken by electron as kinetic energy.
(5) The change in wavelength due to Compton effect is called Compton shift. Compton shift \[{{\lambda }_{f}}-{{\lambda }_{i}}=\Delta \lambda =\frac{h}{{{m}_{0}}c}(1-\cos \varphi )\]
If \[\phi ={{0}^{o}},\,\,\Delta \lambda =0\]
\[\phi ={{90}^{o}},\,\,\Delta \lambda =\frac{h}{{{m}_{0}}c}=0.24\,nm\] \[\varphi ={{180}^{o}},\,\Delta \lambda =\frac{2h}{{{m}_{0}}c}=\,0.48nm\] (called Compton wave length)
(1) \[h\nu =h{{\nu }_{0}}+{{K}_{\max }}\] and \[{{K}_{\max }}=e{{V}_{0}}\]
(2) \[{{K}_{\max }}=e{{V}_{0}}=h(\nu -{{\nu }_{0}})\]\[\Rightarrow \]\[\frac{1}{2}mv_{\max }^{2}=h(\nu -{{\nu }_{0}})\]
(3) \[{{v}_{\max }}=\sqrt{\frac{2h(\nu -{{\nu }_{0}})}{m}}\]
(4) \[{{K}_{\max }}=\frac{1}{2}mv_{\max }^{2}=e{{V}_{0}}=hc\,\left( \frac{1}{\lambda }-\frac{1}{{{\lambda }_{0}}} \right)=hc\,\left( \frac{{{\lambda }_{0}}-\lambda }{\lambda {{\lambda }_{0}}} \right)\]
(5) \[{{v}_{\max }}=\sqrt{\frac{2hc}{m}\frac{\left( {{\lambda }_{0}}-\lambda \right)}{\lambda {{\lambda }_{0}}}}\]
(6) \[{{V}_{0}}=\frac{h}{e}(\nu -{{\nu }_{0}})=\frac{hc}{e}\left( \frac{1}{\lambda }-\frac{1}{{{\lambda }_{0}}} \right)=12375\,\left( \frac{1}{\lambda }-\frac{1}{{{\lambda }_{0}}} \right)\]
(1) Effect of intensity : If the intensity of light is increased (while it's frequency is kept the same) the current levels off at a higher value, showing that more electrons are being emitted per unit time. But the stopping potential \[{{V}_{0}}\] doesn't change i.e.
Intensity \[\propto \] no. of incident photon \[\propto \] no. of emitted photoelectron per time \[\propto \] photo current
(2) Effect of frequency : If frequency of incident light increases, (keeping intensity is constant) stopping potential increases but their is no change in photoelectric current
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