Question

The uncertainty in the position of a ball weighing 50 g is $\pm {10}^{-4}mm$. Calculate the minimum uncertainty in its velocity in m/s

• A. $2.63\times {10}^{-29}m/s$
• B. $1.05\times {10}^{-26}m/s$
• C. $4.5\times {10}^{-30}m/s$
• D. $7.55\times {10}^{-29}m/s$

Answer 1

The correct option is B $1.05\times {10}^{-26}m/s$

From Heisenberg’s uncertainty principle,
Δx. Δp ≥ $\frac{h}{4\pi }$
Since, p = m.v,
Δp = m.(Δv) [Because mass is fixed]

Therefore,
$\Rightarrow \Delta x.m.\left(\Delta v\right)\ge \frac{h}{4\pi }$
where, h = Planck’s constant
Δx = Uncertainty in position = ${10}^{-4}mm$ = ${10}^{-7}m$
Δv = Uncertainty in velocity
m = Mass of the ball in kilograms = $50\times {10}^{-3}kg$
So, $\Delta v\ge \frac{h}{4\pi .m.\Delta x}....\left(1\right)$

Putting up the values in eq. (1)
$\Delta v\ge \frac{6.62\times {10}^{-34}}{4\times 3.14\times 50\times {10}^{-3}\times {10}^{-7}}\Delta v\ge 1.05\times {10}^{-26}m/s$