Question

A base ball of mass `200 g` is moving with velocity of `3 xx 10^(3) cm s^(-1)` .If we can locte the base ball with an error equal to the magnitude of the wavelength of the light used `(5000 Å)` how wil the uncertainty in momentum be used with the total momentum of the base ball?

Answer 1

Momentum of the basic ball = mv
`p = 200 xx 10 = 6 xx 10^(5) cm g s^(-1)`
`Delta p Delta x ge h//4pi`
`Delta p ge (h)/(4pi Delta x) ( Deltax = 5000 xx 10^(-8) = 5 xx 10^(-5))`
` (DeltaP)/Pge (6.626 xx 10^(-27))/(4 xx 3.14 xx 5 xx 10^(-5))`
`( Delta p)/(p) ge (6.626 xx 10^(-27))/(4 xx 3.14 xx 5 xx 10^(-5))`
`= 1.756 xx 10^(-29)`