Question

The uncertainty in position and velocity of a particle are ${{10}^{-10}}m$and$5.27\times {{10}^{-24}}m{{s}^{-1}}$respectively. Calculate the mass of the particle. ($h=6.625\times {{10}^{-34}}J-s$)

Answer 1

HINT: We can solve this problem by using Heisenberg’s Uncertainty Principle which gives us a relation between the position and momentum. Here, we can convert momentum in the terms of mass and velocity and put the respective values to get the answer in kilograms.

COMPLETE STEP BY STEP SOLUTION: Heisenberg’s Uncertainty Principle gives us a relation between the uncertainty in position and momentum.
It states that we cannot measure the position and the momentum of a particle simultaneously, regardless of the quantum mechanical state the particle is in.
Mathematically, it gives the relation-
     \[\Delta x\cdot \Delta p=\dfrac{\hbar }{2}\]
Where, \[\Delta x\]is the uncertainty in position and
\[\Delta p\]is the uncertainty in momentum and
\[\hbar \]is the reduced Planck’s constant.
The reduced Planck’s constant is given as-\[\hbar =\dfrac{h}{2\pi }\],
Where, h is the Planck’s constant and its value is given as- $h=6.625\times {{10}^{-34}}J-s$
In the question,
We are given the uncertainty of position,$\Delta x$=${{10}^{-10}}m$ and
we are given the uncertainty in velocity,$\Delta v$=$5.27\times {{10}^{-24}}m{{s}^{-1}}$
As we know, momentum is the product of mass and velocity,
Therefore, we can write $\Delta p$as-
$\Delta p=m\Delta v$
Or, putting the value of momentum in the Heisenberg’s equation, we get
\[\Delta x\cdot m\Delta v=\hbar \]
Or,\[\Delta x\cdot m\Delta v=\dfrac{h}{4\pi }\]
Therefore, putting the values of uncertainty of position, velocity and Planck’s constant, we will get-
${{10}^{-10}}m\times m\times 5.27\times {{10}^{-24}}m{{s}^{-1}}=\dfrac{6.625\times {{10}^{-34}}J-s}{4\pi }$
Or,$m=\dfrac{6.625\times {{10}^{-34}}J-s}{4\times 3.145\times 5.27\times {{10}^{-24}}m{{s}^{-1}}\times {{10}^{-10}}m}=0.09992J{{s}^{2}}{{m}^{-2}}$=0.09992
We know, $J{{s}^{2}}=kg{{m}^{2}}$
Therefore, the unit conversion can be written as
$J{{s}^{2}}{{m}^{-2}}=kg{{m}^{2}}{{m}^{-2}}=kg$
Therefore, the mass of the particle is 0.09992kg.
However, this principle is only valid for microscopic moving particles as for larger particles the uncertainty in mass and velocity is negligible due to their larger size but for smaller, microscopic particles as they are very small masses thus, it is difficult to find their accurate position without any uncertainty in velocity.

ADDITIONAL INFORMATION: The same principle can also be used written in the terms of time and energy by using the definition of velocity and it states that it is not possible to measure the energy of a particle accurately in a given finite time.

NOTE: It is important to remember the relation between uncertainty of position and uncertainty of momentum to solve this problem. Units should be written always or else the answer might come out to be faulty.