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Find the momentum of a particle whose de Broglie wavelengh is 1Å. |
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Answer» Solution :de BROGLIE wave length, `lambda=1Å=1.0xx10^(-10) m` Momentum of a particle, P = ? `lambda=(H)/(P) or P=(h)/(lambda)=(6.625xx10^(-34)KGM^(2)s^(-1))/(1.0xx10^(-10)m)` Momentum of a particle is `6.625 xx 10^(-24) kg ms^(-1)` |
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