Question

calculate the uncertainty in the velocity of a cricket ball of mass 0.15 kg if the uncertainty in its position is of the order of 100 pm.??​

Answer 1

Answer:

Use uncertainty principle,

e.g., ∆x.m∆v = h/4π

Here, ∆x is uncertainty in position

∆v is uncertainty in velocity.

Here, m = 0.15 kg , ∆x = 1 A° = 10⁻¹⁰ m

h = 6.626 × 10⁻³⁴ J.s

Now, ∆v = h/4πm∆x

= 6.626 × 10⁻³⁴/(4 × 3.14 × 0.15 × 10⁻¹⁰)

= 6.626 × 10⁻²⁴/(0.6 × 3.14)

= 3.51 × 10⁻²⁴ m/s

Hence, uncertainty in velocity = 3.51 × 10⁻²⁴ m/s

Answer 2

Answer:

$\displaystyle{\Delta v\geq3.524\times10^{-24}}$

Step-by-step explanation:

Given :  

Mass of cricket ball = 0.15 kg

Uncertainty position  = 100 pm

In term of meter :  

Uncertainty position  =  $\displaystyle{100\times10^{-12} \ m \ or \ 1.0\times10^{-10} \ m}$

We have  Uncertainty  formula :  

 $\displaystyle{\Delta x \ m \ \Delta v\geq\dfrac{h}{4\pi} }\\\\\\\displaystyle{\Delta v\geq \dfrac{h}{4\pi\Delta x \ m}}$

Putting values now :  

$\displaystyle{\Delta v\geq \dfrac{6.626\times10^{-34}}{4\times3.14\times10^{-10}\times0.15}}\\\\\\\displaystyle{\Delta v\geq \dfrac{6.626\times10^{-24}}{1.88}}\\\\\\\displaystyle{\Delta v\geq3.524\times10^{-24}}$

Hence we get answer .