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Question: A proton is accelerated to one-tenth of the velocity of light. If its velocity can be measured with ...

A proton is accelerated to one-tenth of the velocity of light. If its velocity can be measured with a precision±1%\pm 1\%, then its uncertainty in position is:
A) 1.05 X 10 - 13m{\text{1}}{\text{.05 X 1}}{{\text{0}}^{{\text{ - 13}}}}{\text{m}}
B) 1.13 X 10 - 13m{\text{1}}{\text{.13 X 1}}{{\text{0}}^{{\text{ - 13}}}}{\text{m}}
C) 1.15 X 10 - 13m{\text{1}}{\text{.15 X 1}}{{\text{0}}^{{\text{ - 13}}}}{\text{m}}
D) 1.25 X 10 - 13m{\text{1}}{\text{.25 X 1}}{{\text{0}}^{{\text{ - 13}}}}{\text{m}}

Explanation

Solution

In this question we will use uncertainty principle to find the answer which is as follows: Δx.ΔV=h4πm\Delta x.\Delta V = \dfrac{h}{{4\pi m}}here, Δx\Delta x=represents Uncertainty in position, ΔV\Delta V= represents Uncertainty in velocity, hh= represents Planck's constant =6.626 X 1034 Js6.626{\text{ X }}{10^{ - 34}}{\text{ Js}}, mm= represents Mass of proton =1.673 X 1027 Kg1.673{\text{ X }}{10^{ - 27}}{\text{ Kg}}.

Complete step by step answer:
To proceed with the question, let us first find the velocity and the uncertainty in velocity
Velocity of proton, V=C10=3X10810=3X107m/sV = \dfrac{C}{{10}} = \dfrac{{3 X {{10}^8}}}{{10}} = 3 X {10^7}m/s
Uncertainty in velocity, \Delta V = 1{\text{% of }}V = \dfrac{1}{{100}}{\text{ X }}3{\text{ X }}{10^7} = 3{\text{ X }}{10^5}{\text{m/s}}
From Heisenberg uncertainty principle we know that the product of uncertainty in position and uncertainty in velocity is given by the formula,
Δx.ΔV=h4πm\Delta x.\Delta V = \dfrac{h}{{4\pi m}}where,
Δx\Delta x= Uncertainty in position
ΔV\Delta V= Uncertainty in velocity
hh= Planck's constant = 6.626 X 1034 Js6.626{\text{ X }}{10^{ - 34}}{\text{ Js}}
mm= Mass of proton = 1.673 X 1027 Kg1.673{\text{ X }}{10^{ - 27}}{\text{ Kg}}
Therefore, the uncertainty in position is given by
Δx=h4πmΔV\Delta x = \dfrac{h}{{4\pi m\Delta V}}
Substituting the values calculated above in the above equation we get,
Δx=6.626X1034(4X3.14X1.673X1027X3X105)\Delta x = \dfrac{{6.626 X {{10}^{ - 34}}}}{{(4 X 3.14 X 1.673 X {{10}^{ - 27}} X 3 X {{10}^5})}}
Δx=1.05X1013 m\Rightarrow \Delta x = 1.05 X {10^{ - 13}}{\text{ m}}
Hence, uncertainty in position is1.05X1013 m1.05 X {10^{ - 13}}{\text{ m}}.
Therefore, A is the correct option.

Note:
Now we will study three principles for the atomic structure which are useful during the filling of the orbitals according to the electronic configuration of the atoms.
-The aufbau principle also known as aufbau rule, given by Aufbau states that in the feeling of the electrons first electrons occupy the lowest energy level and then the higher energy levels.
-Hund's rule: It states that in every orbital first it will occupy single or unpaired electrons and then pairing of electrons will take place.
Pauli's Exclusion Principle states that the electrons in the orbital will have opposite spin.
-According to Heisenberg's uncertainty principle it is not possible to calculate the position and the velocity of the atom with precision. It provides only the rough data for the position and the velocity not the exact correct data.