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If uncertainities in the measurement of position and momentum of a microscopic object of mass ' $m$ are equal, then the uncertainty in the measurement of velocity is given by the expression
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Verified Answer
The correct answer is:
$\sqrt{\frac{h}{4 \pi}} \times \frac{1}{m}$
According to Heisenberg's uncertainity principle.
It is impossible to determine simultaneously the position and momentum of moving particle thus.
$$
\Delta x \times \Delta p \geq \frac{h}{4 \pi}
$$
But, $\quad p=m \times v$
$$
\begin{array}{rlrl}
& \therefore & \Delta v^2 & =\frac{h}{m^2 4 \pi} \\
\text { or } & \Delta v & =\sqrt{\frac{h}{4 \pi}} \times \frac{1}{m}
\end{array}
$$
It is impossible to determine simultaneously the position and momentum of moving particle thus.
$$
\Delta x \times \Delta p \geq \frac{h}{4 \pi}
$$
But, $\quad p=m \times v$
$$
\begin{array}{rlrl}
& \therefore & \Delta v^2 & =\frac{h}{m^2 4 \pi} \\
\text { or } & \Delta v & =\sqrt{\frac{h}{4 \pi}} \times \frac{1}{m}
\end{array}
$$
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