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?

Answer

`1.75xx10^(-29)`

Explanation

`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)`