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A base ball of mass `200 g` is moving with velocity of `3 xx 10^(3) cm s^(-1)`. If we can locate the base ball with an error equal to the magnitude of the wavelength of the light used `(5000 Å)`. How will the uncertainty in momentum be used with the total momentum of the base ball? |
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Answer» Correct Answer - `1.75xx10^(-29)` `DeltaV =30xx10^(2)cm//sec` `lambda=5000Å" "m=200g` `lambda=(h)/(mV)" "500=(h)/(mxxV)` `P=mV=(500)/(6.626xx10^(-26))=30xx10^(2)xx200` `=1.75xx10^(-29)` |
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