1.

Find the accelerating potential of the electron, when its de-Broglie wave length is 1Å. Data: lambda=Å, V = ?

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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