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A photon of wavelength 1.4 Å collides with an electron. After collision, the wavelength of the photon is found to be 2.0 Å. Calculate the energy of the scattered electron.

Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/energy-15288" style="font-weight:bold;" target="_blank" title="Click to know more about ENERGY">ENERGY</a> of the photon before <a href="https://interviewquestions.tuteehub.com/tag/collison-2031742" style="font-weight:bold;" target="_blank" title="Click to know more about COLLISON">COLLISON</a> `= hv = (<a href="https://interviewquestions.tuteehub.com/tag/hc-1016346" style="font-weight:bold;" target="_blank" title="Click to know more about HC">HC</a>)/(lamda)` <br/> Energy of the photon after collision `= hv' = (hc)/(lamda')` <br/> Difference of energy is imparted to the electron. <br/> Hence, energy of the scattered electron `= (hc)/(lamda) - (hc)/(lamda') = hc ((1)/(lamda)- (1)/(lamda'))` <br/> `= (6.626 xx 10^(-34) <a href="https://interviewquestions.tuteehub.com/tag/j-520843" style="font-weight:bold;" target="_blank" title="Click to know more about J">J</a> s) (3 xx 10^(<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a>) m s^(-1)) xx((1)/(1.4 xx 10^(-10)m) - (1)/(2.0 xx 10^(-10) m))` <br/> `= (6.626 xx 10^(-34)Js) (3 xx 10^(8) ms^(-1)) xx ((2.0 - 1.4) xx 10^(-10) m^(-1))/((1.4 xx 10^(-10)) (2.0 xx 10^(-10))) = 4.26 xx 10^(-16) J`</body></html>


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