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Calculate the momentum of a particle which has de Broglie wavelength of `2.5xx10^(-10)m`. |
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Answer» Correct Answer - `2.65xx10^(-24) "kg " m s^(-1)` According to de Broglie equation, ` " " lambda(h)/(p) or p(h)/(lambda)` `p=((6.626xx10^(-34) "kg "m^(2)s^(-1)))/((2.5xx10^(-10)m))=2.65xx10^(-24) "kg " m s^(-1)`. |
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