Question

If uncertainty in the position of an electron is zero, the uncertainty in its momentum would be: -
(A). Zero
(B). $ < \dfrac{h}{{4\pi }}$
(C). $ > \dfrac{h}{{4\pi }}$
(D). infinite

Answer 1

Hint: The uncertainty principle was proposed by Heisenberg in $1927$ . It was one of the laws which failed all the classical theories of atomic structure and led to the quantum structure of an atom. The other laws are Dual nature of matter (by de-Broglie in $1924$ ), the black body radiation and the photoelectric effect.

Complete answer:
Heisenberg was a German Physicist.
According to Heisenberg's uncertainty principle, we cannot measure the position and the momentum of any moving particle with accuracy. Thus we have $\Delta x \times \Delta {p_x} \geqslant \dfrac{h}{{4\pi }}$
Where $\Delta x$ is uncertainty in the position, $\Delta {p_x}$ is the uncertainty in the momentum and $h$ is planck's constant. If $\Delta x \to 0$ then $\Delta {p_x} \to \infty $ .
Thus the correct answer is (D). infinite
As we know that momentum is the product of mass and velocity, thus iff we substitute this relation to the above equation the modified Heisenberg equation is $\Delta x \times m\Delta {v_x} \geqslant \dfrac{h}{{4\pi }}$ .
On rearranging it we get $\Delta x \times \Delta {v_x} \geqslant \dfrac{h}{{4\pi m}}$
Where $\Delta {v_x}$ is the velocity of the microscopic moving particle and $m$ is the mass of the microscopic moving particle.
Thus we can also say that we cannot measure the velocity and the position of any microscopic moving particle with high accuracy.

Note:
The statement of Heisenberg's uncertainty principle is as follows: “The position and momentum of any microscopic moving particle cannot be determined simultaneously with accuracy or certainty. If an attempt is made to measure any one of these quantities with higher accuracy, the other becomes less accurate.” To his uncertainty principle, he was awarded with the nobel prize in $1932$ under the branch of atomic physics.