Given below are two statements :

Statement (I): It is impossible to specify simultaneously with arbitrary precision, both the linear momentum and the position of a particle.

Statement (II) : If the uncertainty in the measurement of position and uncertainty in measurement of momentum are equal for an electron, then the uncertainty in the measurement of velocity is $\geqslant \sqrt{\frac{h}{\pi}} \times \frac{1}{2 m}$.

In the light of the above statements, choose the correct answer from the options given below :

<p>Statement I : Correct</p> <p>This statement is describing the uncertainty principle.</p> <p>It states that it is impossible to measure both the position and the momentum of an object.</p> <p>Statement II : Correct.</p> <p>If the uncertainty in the measurement of position ($\Delta x$) and the uncertainty in the measurement of momentum ($\Delta p$) are equal for an electron, then the uncertainty in the measurement of velocity ($\Delta v$) is greater than or equal to $\sqrt{\frac{h}{\pi}}\frac{1}{2m}$</p> <p>This is derived from the Heisenberg uncertainty principle, $\Delta x\,.\,\Delta {p_x} \ge {h \over {4\pi }}$</p> <p>Given, $\Delta x = \Delta p$</p> <p>Substitute this in $\Delta x\,.\,\Delta p \ge {h \over {4\pi }}$ as</p> <p>$\Delta p\,.\,\Delta p \ge {h \over {4\pi }}$</p> <p>$\Delta {p^2} \ge {h \over {4\pi }}$</p> <p>$\Delta {p^2} \ge {h \over {4\pi }}$</p> <p>$$\left\{ \matrix{ The\,formula\,for\,momentum\,(P)\,is\,P = mV \hfill \cr For\,uncerta{\mathop{\rm int}} y\,\Delta P = m\Delta v \hfill \cr} \right.$$</p> <p>So, $m\Delta v \ge \sqrt {{h \over {4\pi }}}$</p> <p>$\Delta v \ge \sqrt {{h \over {4\pi }}} \times {1 \over m}$</p> <p>$\Delta v \ge {1 \over 2} > \sqrt {{h \over \pi }} > {1 \over m}$</p> <p>$\Delta v \ge \sqrt {{h \over \pi }} {1 \over {2m}}$</p> <p>Both the statements are true (correct). So answer is (2) Both statement I and statement II are true.</p>