Calculate the uncertainty in velocity of a cricket ball of mass 150 g the uncertainty in its position is of the order of 1 mathring - A - left- h-6-6times - 10 - -34 -kg- m - 2 - s - -1 - right- -

Question

Calculate the uncertainty in velocity of a cricket ball of mass 150 g the uncertainty in its position is of the order of 1 mathring - A -  left- h-6-6times - 10 - -34 -kg- m - 2 - s - -1 - right- -

Toppr Admin · Accepted Answer

M-150-xD7-10-x2212-3kg-x394-x-10-x2212-10A0 h-planck-apos-s constant-xA0-Using Heisenburg Uncertainity Principle-x394-x-m-x394-v-x2265-h4-x3C0-x394-v-h4-x3C0-m-x394-x-x394-v-6-626-xD7-10-x2212-344-xD7-3-14-xD7-10-x2212-10-xD7-0-15-3-5-xD7-10-x2212-24ms-x2212-1

Calculate the uncertainty in velocity of a cricket ball of mass $150 g$ if the uncertainty in its position is of the order of $1 ˚ A$ $(h = 6.6 \times 10^{- 34} k g m^{2} s^{- 1})$ .

$m = 150 \times 10^{- 3} k g Δ x = 10^{- 10} A^{0}$ h=planck's constant
Using Heisenburg Uncertainity Principle
$Δ x . m Δ v \geq \frac{h}{4 π} Δ v = \frac{h}{4 π m Δ x} Δ v = \frac{6.626 \times 10^{- 34}}{4 \times 3.14 \times 10^{- 10} \times 0.15} = 3.5 \times 10^{- 24} {m s}^{- 1}$

Calculate the uncertainty in velocity of a cricket ball of mass 150g if the uncertainty in its position is of the order of 1˚A (h=6.6×10−34kgm2s−1).

m=150×10−3kgΔx=10−10A0 h=planck's constant Using Heisenburg Uncertainity PrincipleΔx.mΔv≥h4πΔv=h4πmΔxΔv=6.626×10−344×3.14×10−10×0.15=3.5×10−24ms−1

Calculate the uncertainty in velocity of a cricket ball of mass $150 g$ if the uncertainty in its position is of the order of $1 ˚ A$ $(h = 6.6 \times 10^{- 34} k g m^{2} s^{- 1})$ .

$m = 150 \times 10^{- 3} k g Δ x = 10^{- 10} A^{0}$ h=planck's constant
Using Heisenburg Uncertainity Principle
$Δ x . m Δ v \geq \frac{h}{4 π} Δ v = \frac{h}{4 π m Δ x} Δ v = \frac{6.626 \times 10^{- 34}}{4 \times 3.14 \times 10^{- 10} \times 0.15} = 3.5 \times 10^{- 24} {m s}^{- 1}$