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
PLS GIVE THE ANSWER
- 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
- 0.7 because 70% of energy is converted to photon.Final answer will be in picometer (pm)
- 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)
- 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)
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