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AN XRAY TUBE OPERATES AT 20KV SUPPOSE THE ELECTRON CONVERTS 70% OF ITS ENERGY INTO A PHOTON AT EACH COLLISION . FIND THE LOWEST THREE WAVELENGHTS EMMITED EMMITED FROM THE TUBE NEGLECT THE ENERGY IMPARED TO THE ATOM WITH WHICH THE ELECTRON COLLIDES
 
 
 
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Asked by sudheerkapoor67 9th October 2017, 6:24 PM
Answered by Expert
Answer:
  1. For the first wavelength the a photon of 70% energy is emmited. For the next wavelength 30% of photon energy is used for emiisoin of next wavelength when it collide internally with other electrons. similarly for third wavelength
  2. begin mathsize 12px style straight lambda subscript 1 equals hc over straight E equals fraction numerator 4.135 straight x 10 to the power of negative 15 end exponent straight x 3 straight x 10 to the power of 8 over denominator 0.7 straight x 20 straight x 10 cubed end fraction end style  0.7 because 70% of energy is converted to photon.Final answer will be in picometer (pm)
  3. Error converting from MathML to accessible text. 0.3x0.7 is because after collision of photon inside the atom the energy left is 30% of initial energy .Final answer will be in picometer (pm)
  4. Error converting from MathML to accessible text. 0.3x0.3x0.7 because of each collision of photon inside the atom the energy left is 30% . Final answer will be in picometer (pm)
Answered by Expert 27th November 2017, 6:27 PM
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