Question

The measurement of the electron position is associated with an uncertainty in electron velocity, momentum, which is equal to $1 \times {10^{ - 18}}$ g cm/s. The uncertainty in electron velocity is, (mass of an electron is $9 \times {10^{ - 28}}$ g)
(A) $1 \times {10^{11}}$ cm/s
(B) $1 \times {10^9}$ cm/s
(C) $1 \times {10^6}$ cm/s
(D) $1 \times {10^5}$ cm/s

Answer 1

Hint : Heisenberg’s uncertainty principle states that it is impossible to measure or calculate exactly, both the position and the momentum of an object. The uncertainty principle is expressed in terms of a particle's momentum and position. The momentum of a particle is equal to the product of its mass times its velocity. Uncertainty in position is the accuracy of the measurement. Thus, the smallest uncertainty in momentum can be calculated. Once the uncertainty in momentum is found, the uncertainty in velocity can be found.

Complete Step By Step Answer:
As per the Heisenberg's uncertainty principle equation,
momentum is given by,
$\Rightarrow$ $\Delta p = m\Delta v$
Where, $\Delta p$ is the error in the measurement of momentum, $\Delta v$ is the error in the measurement of velocity and assuming mass remaining constant during the experiment.
$\Delta v = \dfrac{{1 \times {{10}^{ - 18}}}}{{9 \times {{10}^{ - 28}}}}$
$\Delta v = 0.92 \times {10^{10}}$ cm/s
$\Rightarrow$ $\Delta v \approx 1 \times {10^9}$ cm/s
Hence the correct option is 2) $1 \times {10^9}$ cm/s.

Note :
Heisenberg’s uncertainty principle is a fundamental theory that explains why it is impossible to measure more than one quantum variable simultaneously. Another implication of the uncertainty principle is that it is impossible to accurately measure the energy of a system in some finite amount of time.
If, $\Delta x$ is the error in position measurement and $\Delta p$ is the error in the measurement of momentum, then
$\Delta x \times \Delta p \geqslant \dfrac{h}{{4\pi }}$
Since momentum, $p = mv$
Then formula can be alternatively written as
$\Delta x \times \Delta \left( {mv} \right) \geqslant \dfrac{h}{{4\pi }}$ or $\Delta x \times \Delta m \times \Delta v \geqslant \dfrac{h}{{4\pi m}}$ .