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The K.E. of the electron is E, when the incident wavelength is lamda. To increase the K.E. of the electron to 2E, the incident wavelength must be : |
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Answer» `(HC)/(Elamda-hc)` `or (hc)/(lambda)-E=(hc)/(x)-2E,` SOLVING `x=(hc lambda)/(lambda+hc)` |
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