Answer Show  Hint: We know that Heisenberg’s uncertainty principle states that it is not possible to measure simultaneously the position and momentum of a microscopic particle with absolute accuracy or certainty. Complete step by step answer: Let’s understand the mathematical expression of Heisenberg’s uncertainty principle. Mathematically, the product of uncertainty in position and uncertainty in momentum of a microscopic particle is always constant and is equal to or greater than $dfrac{h}{4\pi}$.$\Delta x.\Delta p \geqslant \dfrac{h}{4\pi} $…… (1)Where, $\Delta x$ is uncertainty in measuring exact position and $\Delta p$ is uncertainty in measuring exact momentum.From equation (1), it is evident that If $\Delta x$ is very small, the position of microscopic particles can be measured accurately, But $\Delta p$ will be very large which means that momentum or velocity of the particle cannot be measured with accuracy.We know that $p = mv$ , where p is momentum, m is mass and v is velocity So, we replace $\Delta p$ by $m\Delta v$. So, equation (1) becomes,$\Delta x.m\Delta v \geqslant \dfrac{h}{4\pi}$Now, we have to have to rearrange the above equation to calculate $\Delta x$.\[\Delta v \geqslant \dfrac{h}{{4\pi m\Delta x}}\]…… (2)The mass of electron is $9.11 \times {10^{ - 31}}\,{\text{kg}}$, the value of Planck’s constant is $6.626 \times {10^{ - 34}}\,{{\text{m}}^{\text{2}}}\,{\text{kg/s}}$, the value of $\Delta x$Is $4.782 \times {10^{ - 3}}\,{\text{m}}$Now, we have to put all the above values in equation (2).\[ \Rightarrow \Delta v = \dfrac{{6.626 \times {{10}^{ - 34}}\,{{\text{m}}^{\text{2}}}\,{\text{kg/s}}}}{{4 \times 3.14 \times 9.11 \times {{10}^{ - 31}}\,{\text{kg}} \times 4.782 \times {{10}^{ - 3}}\,{\text{m}}}} = \dfrac{{6.626}}{{574.16}} = 0.012\,m/s\]Therefore, the uncertainty of velocity of the electron is 0.012 m/s. Note: It is to be noted that the uncertainty principle has no impact on our daily life. It applies to the moving microscopic particles (protons, neutrons and electrons) which we can not see with our eye. It has no influence on the moving macro or semi micro objects which we actually see or observe. Thus, the Heisenberg’s uncertainty principle has no impact on our daily life.
This means that the uncertainty in the velocity of an electron is 193x greater than the speed of light, not the velocity of the electron. Logically this means that upon measuring the position of an electron to such high precision a simultaneous measurement of its velocity must be proportionately imprecise.Fundamentally this means that if we were capable of calibrating a laser to measure the position of an electron to that precision then the laser would impart a massive amount of kinetic energy upon the electron. Arguably this would be enough energy to accelerate the electron to relativistic speeds. DiscussionSuppose this were an experiment Electron PositionThe position of an electron was measured to be 1067 femtometers ($1067 \times 10^{-15} m$) with an uncertainty of 1 femtometer ($10^{−15}m$). Assuming our experiment was well performed and we have confidence in this measurement, we could state with confidence and certainty that the electron's position is $1067 \pm 1$ femtometers. Therefore we can say with confidence that the actual position could be 1066-1068 femtometers. One would say this is an extremely precise measurement. For the benefit of readers in introductory physics or engineering classes, this would qualitatively sum up to a percent error of 0.0937% Electron VelocityNow let's invert the problem. The velocity of the electron was measured to be 6,843,651 m/s with an uncertainty of $5.79 \times 10^{10}$. Since the uncertainty in the measurement is far greater than the measurement, we know a priori that the measurement should probably not be trusted. We go ahead and claim that the electron's velocity is $6.843651 \times 10^{6} m/s \pm 10^{15} m/s $; but, that would mean that the range of possible values for the velocity is +1,000,000,006,843,651 m/s to -999,999,993,156,349 m/s. In SI units: 1 Pm/s to -1fm/s. With such a massive range of values, we can't say with any certainty what the velocity of the electron is. Therefore the measurement is extremely imprecise The Uncertainty Principle.Fundamentally, the physics behind the uncertainty principle are founded upon how we measure things at the atomic scale. If we need to know the position of particle A, shoot laser B or particle B at it and measure the response of B. With a laser our measurement error in the position is directly tied to the wavelength of our laser. The smaller the wavelength, the better our measurement. With a particle it'd be tied to speed. This is great until we realize that by impacting Particle A with a high velocity particle B or a highly energetic laser (wavelength is inversely proportional to energy remember?) we are imparting energy to it. Basic kinematics with momentum or kinetic energy tells us that the more energetic our particle or laser, the more likely Particle A will absorb that energy and fly in some direction at blazing fast velocities. The end result is an equal imprecision in any measurement of the particle's velocity when we measure its position. Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses No worries! We‘ve got your back. Try BYJU‘S free classes today! No worries! We‘ve got your back. Try BYJU‘S free classes today! No worries! We‘ve got your back. Try BYJU‘S free classes today! Open in App Suggest Corrections |
What will be the uncertainty in velocity of an electron if uncertainty in its position is 1a?
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