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Question: A microscope using suitable photons is employed to locate an electron in an atom within a distance o...

A microscope using suitable photons is employed to locate an electron in an atom within a distance of 0.1A.0.1A. What is the uncertainty involved in the measurement of its velocity?
( h=6.626 ×10−34 Jsh=6.626~\times {{10}^{-34}}~Js , mass of electron =9.1 ×10−31 kg =9.1~\times {{10}^{-31}}~kg~ )

Explanation

Solution

Hint : The uncertainty principle says that the position and momentum of the object cannot be measured precisely at same time duration. The uncertainty in velocity is given by the formula which says that momentum is equal to mass multiplied by the velocity.

Complete Step By Step Answer:
It is given that the measurement of position and momentum of an electron are equal. So, the equation is given as; Δx=Δp\Delta x=\Delta p where , Δx\Delta x is uncertainty in position; Δp\Delta p is uncertainty is momentum.
The uncertainty principle says that the position of the object and velocity of the object cannot be determined at the same time. The uncertainty principle is given by the equation as;
Δx×Δp≥h4π\Delta x\times \Delta p\ge \dfrac{h}{4\pi }
Where, h is the Planck’s constant as both the position and momentum are equal, then the equation can be written as; (Δp)2≥h4π{{\left( \Delta p \right)}^{2}}\ge \dfrac{h}{4\pi }
⇒(Δp)≥(h4π)2\Rightarrow \left( \Delta p \right)\ge {{\left( \dfrac{h}{4\pi } \right)}^{2}}
According to Heisenberg’s Uncertainty principle: del(X)del(P)=h2pidel\left( X \right)del\left( P \right)=\dfrac{h}{2pi} and we have m=9.11×10−31kgm=9.11\times {{10}^{-31}}kg and h=6.626×10−34Jsh=6.626\times {{10}^{-34}}Js
Now just we have substitute the given thing in formula;
ΔV≥h4πmΔx\Delta V\ge \dfrac{h}{4\pi m\Delta x}
⇒ΔV≥6.626×10−344×3.14×9.11×10−31(0.1×10−10)\Rightarrow \Delta V\ge \dfrac{6.626\times {{10}^{-34}}}{4\times 3.14\times 9.11\times {{10}^{-31}}\left( 0.1\times {{10}^{-10}} \right)} we get;
⇒ΔV≥0.579×107ms−1\Rightarrow \Delta V\ge 0.579\times {{10}^{7}}m{{s}^{-1}}
⇒ΔV≥5.79×106ms−1\Rightarrow \Delta V\ge 5.79\times {{10}^{6}}m{{s}^{-1}}
Therefore, uncertainty in the measurement of velocity is 5.79×106ms−15.79\times {{10}^{6}}m{{s}^{-1}} .

Note :
The uncertainty principle is also known as the Heisenberg uncertainty principle or indeterminacy principle. The functions which are non-commutative follow this principle. So the uncertainty in momentum is given by mass multiplied by uncertainty in velocity. The equation to calculate uncertainty in momentum.