Question

Can you locate the electron within 0.005 nm?

Answer 1

The electron is located in the orbitals around the nucleus. But, since electrons are always in motion so, it is impossible to tell the exact location. It is only possible to find the probability of an electron at a certain place within an atom.

Complete answer:
Let us first see the concept behind solving this question, the Heisenberg uncertainty principle explains the restrictions on the accuracy of simultaneous measurement of position and momentum. More is the precision in measurement of position, less accurate momentum will be and vice-versa. It also states that it is impossible to measure both the position and the momentum of an object exactly. It is based on the wave-particle duality of matter.
Let us now see its formula,

Let be uncertainty in position and be uncertainty in momentum, then, $\vartriangle \text{x}\times \vartriangle \text{p}\ge \dfrac{\text{h}}{\text{4}\pi }$, we know that momentum is equal to mass multiplied to velocity. $\vartriangle \text{p = m }\times \vartriangle \text{v}$, the formula is transformed to$\vartriangle \text{x}\times \vartriangle \text{v}\ge \dfrac{\text{h}}{\text{4}\pi \text{m}}$.

Step (1)- To solve the numerical, we need to first write what is given to us; $\text{m = 9}\text{.11}\times \text{1}{{\text{0}}^{-31}}\text{kg}$, $\text{h = 6}\text{.626}\times \text{1}{{\text{0}}^{-34}}\text{J-s}$ and$\vartriangle \text{x = 0}\text{.005 }\times \text{1}{{\text{0}}^{-9}}\text{m}$.

Step (2)- Apply the formula, $5\times {{10}^{-12}}\times \vartriangle \text{v }\ge \dfrac{6.626\times {{10}^{-34}}}{4\times 3.14\times 9.11\times {{10}^{-31}}}$ then, $\vartriangle \text{v}$ is greater or equal to$\text{1}\text{.15}\times \text{1}{{\text{0}}^{7}}\text{m/s}$. The speed of light is $3\times {{10}^{8}}\text{m/s}$.

Step (3)- Write the final answer $\vartriangle \text{v}\ge \text{1}\text{.15}\times \text{1}{{\text{0}}^{7}}\text{m/s}$, so that electron cannot be located because any quantity cannot have speed more than speed of light.

Note: This principle cannot be applied to macroscopic objects; Heisenberg’s uncertainty principle has a negligible impact on measurements in the macroscopic objects. While measuring the position of any object, there will be a transfer of momentum from the photons to that object. The mass of the photon is much smaller than the mass of any object. Thus, the momentum imparted by the photon to the ball can be neglected.