Question

The measurement of the electron position is associated with an uncertainty in momentum, which is equal to $1 \times {10^{ - 18}}gcm{s^{ - 1}}.$ the uncertainty in electron velocity is (mass of electron is $9 \times {10^{ - 28}}g)$?

Answer 1

According to classical physics the formula for momentum is given by the product of mass and velocity of the object (p=mv). In the question, the mass of the electron is given, uncertainty in position is given to the uncertainty in velocity can be calculated from the momentum ($\Delta p = m\Delta v$) value itself.

Complete solution:

Heisenberg gave the relation for calculating momentum and position of the electron simultaneously, by the following equation $(\Delta x)(\Delta p) \geqslant \dfrac{h}{{4\pi }}or(\Delta x)(\Delta v) \geqslant \dfrac{h}{{4\pi m}}$ where,

$\Delta x$ - Uncertainty associated with the position of the electron

$\Delta {\text{p}}$ - Uncertainty associated with the momentum of the electron

$\Delta v$ -uncertainty associated with the velocity of the electron

$h$- Planck’s constant

$m$- Mass of the electron

We know that momentum is the product of mass and velocity (p=mv). We are going to apply the momentum formula for calculating the velocity of an electron at their uncertainties.so, let us calculate the uncertainty associated with the velocity from $\Delta p$, where $\Delta p = m\Delta v$.

Given that the uncertainty in momentum is$1 \times {10^{ - 18}}gcm{s^1}$.

The mass of an electron is $9 \times {10^{ - 28}}g$

$\Delta p = m\Delta v$

$1 \times {10^{-18}} = 9 \times {10^{ - 28}}\Delta v$

$\Delta v = \dfrac{{{{10}^{28}} \times {{10}^{^{-18}}}}}{9} = 1.11 \times {10^{9}}cm{\kern 1pt} {\kern 1pt} {s^{ - 1}}$

Hence, the uncertainty of an electron velocity is $1.11\times 10^9cm{s^{ - 1}}$.

Additional information:

Heisenberg’s uncertainty principle is one of the theories which explain the dual nature of matter. It was started by Werner Heisenberg in 1927.

It states that “it is impossible to determine the exact position and exact momentum of an electron in an atom simultaneously” and is given by the equation $\Delta x{\kern 1pt} {\kern 1pt} {\kern 1pt} \times {\kern 1pt} {\kern 1pt} {\kern 1pt} \Delta p{\kern 1pt} {\kern 1pt} \geqslant {\kern 1pt} {\kern 1pt} \dfrac{h}{{4\pi }}$

The effect of the Heisenberg uncertainty principle is significant only for the motion of very small objects such as subatomic particles and is negligible for macroscopic objects.

Therefore to find the exact location and speed of microscopic particles like electrons, it is not possible to find out at the same time precisely. Heisenberg’s uncertainty principle helps in determining the uncertainties of position and velocities of electrons by probability but not by precision. It eliminates the trajectory of definite paths (orbits) in an atom but it explains the probability of finding the particles in an atom (orbitals) i.e. quantum mechanical explanation of the atom.

Note: The given question has to be read carefully as it is mentioned that uncertainty is dependent on the position of the electron. We may think of applying Heisenberg’s equation to know the velocity (v). Here the value of uncertainty in momentum (p) is given but not the value of uncertainty in position(x). So Heisenberg’s formula need not be applied to calculate the velocity of the moving electron.