Question

A wet ball weighing $10.1\text{gm}$ has water of $0.1\text{g}$ on it. The ball is moving with a constant velocity with an uncertainty of momentum of ${10}^{-6}\text{kg m/s}$. What will be the ratio of uncertainty in the measurement of the position of the ball, water and electron in the water molecule ? (mass of the electron, $=9.1\times {10}^{-28}\text{g})$

• A. $1:100:1.1\times {10}^{28}$
• B. $1.1:1:1.1\times {10}^{28}$
• C. $10:10:1.1\times {10}^{28}$
• D. $10:100:1.1\times {10}^{28}$

Answer 1

The correct option is A $1:100:1.1\times {10}^{28}$

From Heisenberg’s Uncertainty Principle,

$\Delta x\times \Delta p\ge \frac{h}{4\pi }$

$\Rightarrow \Delta x\times m\Delta v\ge \frac{h}{4\pi }$

$\Rightarrow \Delta x\ge \frac{h}{4\pi m\Delta v}$

Velocity being constant, uncertainty in the measurement of the momentum is associated with the mass of the matter.

$\therefore \Delta x\propto \frac{1}{m}$

$\Rightarrow \Delta x_{ball}:\Delta x_{water}:\Delta x_{electron}=\frac{1}{m_{ball}}:\frac{1}{m_{water}}:\frac{1}{m_{electron}}$

$=\frac{1}{(10.1-0.1)}:\frac{1}{0.1}:\frac{1}{9.1\times {10}^{-28}}=1:100:1.1\times {10}^{28}$

Hence, $\left(\text{A}\right)$ is the correct answer.