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Find the accelerating potential of the electron, when its de-Broglie wave length is 1Å. Data: lambda=Å, V = ? |
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Answer» SOLUTION :`LAMBDA=(12.27)/(sqrtV)` `sqrtV=(12.27xx10^(-10))/(lambda)` `(12.27xx10^(-10))/(1XX10^(-10))` `V=(12.27)^(2)=150.55V` |
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