
In an atom, an electron is moving with a speed of 600 m/s with an accuracy of 0.005%, certainty with which the position of the electron can be located is: ($h =$ $6.6\times {{10}^{-34}}kg{{m}^{2}}{{s}^{-1}}$, mass of electron ${{e}_{m}}=9.1\times {{10}^{-31}}Kg$).
(A) $1.52\times {{10}^{-4}}m$
(B) $5.01\times {{10}^{-3}}m$
(C) $1.92\times {{10}^{-3}}m$
(D) $3.84\times {{10}^{-3}}m$
Answer
561.3k+ views
Hint: The location of electrons in a nucleus is determined by the Heisenberg’s Uncertainty Principle. The accuracy calculations for velocity and position is done by the relation stated by the same.
Complete step by step solution:
Let us know what do we mean by Heisenberg’s Uncertainty Principle and then concentrate on the given illustration,
Heisenberg’s Uncertainty Principle-
It states that it is not possible to measure both position and momentum (or velocity) of a microscopic particle simultaneously with specific accuracy.
Mathematically, the above statement will be expressed as,
\[\Delta x\times \Delta p\ge \dfrac{h}{4\pi }\]
where,
$\Delta x$= uncertainty in position
$\Delta p$= uncertainty in momentum
The sign $\ge $ means that the product of $\Delta x$ and $\Delta p$ can be either greater than or equal to \[\dfrac{h}{4\pi }\]. It can be never less than \[\dfrac{h}{4\pi }\] .
Fact-
If the position of the particle is measured accurately, then there will be more error in the measurement of momentum and vice versa is true.
Now,
As momentum is a product of mass and velocity. Therefore, we can say,
\[\Delta p=m\Delta v\]
mass is constant.
Thus,
\[\Delta x\times \Delta v\ge \dfrac{h}{4\pi m}\]
It means that the position and velocity cannot be calculated simultaneously with accuracy.
Illustration-
Given that,
$h =$ $6.6\times {{10}^{-34}}kg{{m}^{2}}{{s}^{-1}}$
${{e}_{m}}=9.1\times {{10}^{-31}}Kg$
As we know,
\[\Delta x\times \Delta v\ge \dfrac{h}{4\pi m}\]
Thus,
$\Delta v=600\times \dfrac{0.005}{100}=0.03m/\sec $
as accuracy is given as 0.005%.
Now,
$\begin{align}
& \Delta x\ge \dfrac{h}{4m\Delta v\pi } \\
& \Delta x\ge \dfrac{6.6\times {{10}^{-34}}}{4\times 9.1\times {{10}^{-31}}\times 0.03\times 3.14} \\
& \Delta x\ge 1.9248\times {{10}^{-3}}m \\
\end{align}$
Uncertainty in position = $1.92\times {{10}^{-3}}m$
Therefore, option (C) $1.92\times {{10}^{-3}}m$ is correct.
Note: The uncertainty principle is a fundamental limitation of nature as this principle has no significance in our daily lives. Also, while calculating uncertainty for the substances which are invisible to our naked eyes, simultaneous calculations are not possible for velocity and position.
Complete step by step solution:
Let us know what do we mean by Heisenberg’s Uncertainty Principle and then concentrate on the given illustration,
Heisenberg’s Uncertainty Principle-
It states that it is not possible to measure both position and momentum (or velocity) of a microscopic particle simultaneously with specific accuracy.
Mathematically, the above statement will be expressed as,
\[\Delta x\times \Delta p\ge \dfrac{h}{4\pi }\]
where,
$\Delta x$= uncertainty in position
$\Delta p$= uncertainty in momentum
The sign $\ge $ means that the product of $\Delta x$ and $\Delta p$ can be either greater than or equal to \[\dfrac{h}{4\pi }\]. It can be never less than \[\dfrac{h}{4\pi }\] .
Fact-
If the position of the particle is measured accurately, then there will be more error in the measurement of momentum and vice versa is true.
Now,
As momentum is a product of mass and velocity. Therefore, we can say,
\[\Delta p=m\Delta v\]
mass is constant.
Thus,
\[\Delta x\times \Delta v\ge \dfrac{h}{4\pi m}\]
It means that the position and velocity cannot be calculated simultaneously with accuracy.
Illustration-
Given that,
$h =$ $6.6\times {{10}^{-34}}kg{{m}^{2}}{{s}^{-1}}$
${{e}_{m}}=9.1\times {{10}^{-31}}Kg$
As we know,
\[\Delta x\times \Delta v\ge \dfrac{h}{4\pi m}\]
Thus,
$\Delta v=600\times \dfrac{0.005}{100}=0.03m/\sec $
as accuracy is given as 0.005%.
Now,
$\begin{align}
& \Delta x\ge \dfrac{h}{4m\Delta v\pi } \\
& \Delta x\ge \dfrac{6.6\times {{10}^{-34}}}{4\times 9.1\times {{10}^{-31}}\times 0.03\times 3.14} \\
& \Delta x\ge 1.9248\times {{10}^{-3}}m \\
\end{align}$
Uncertainty in position = $1.92\times {{10}^{-3}}m$
Therefore, option (C) $1.92\times {{10}^{-3}}m$ is correct.
Note: The uncertainty principle is a fundamental limitation of nature as this principle has no significance in our daily lives. Also, while calculating uncertainty for the substances which are invisible to our naked eyes, simultaneous calculations are not possible for velocity and position.
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