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Question: If the mass of the electron is \(9.11\, \times {10^{ - 31}}\) kg. Plank's constant is \(6.626 \times...

If the mass of the electron is 9.11 ×10−319.11\, \times {10^{ - 31}} kg. Plank's constant is 6.626× 10−346.626 \times \,{10^{ - 34}} J-s and uncertainty in position is 0.1 Ao{\text{0}}{\text{.1}}\,{{\text{A}}^{\text{o}}}, then uncertainty in velocity is?
A.5.79  ×108 m/s5.79\,\, \times {10^8}\,{\text{m/s}}
B. 5.79  ×105 m/s5.79\,\, \times {10^5}\,{\text{m/s}}
C.5.79  ×106 m/s5.79\,\, \times {10^6}\,{\text{m/s}}
D. 5.79  ×107 m/s5.79\,\, \times {10^7}\,{\text{m/s}}

Explanation

Solution

We should know the Heisenberg uncertainty principle to answer this question. The Heisenberg uncertainty principle relates the uncertainty in velocity, and position with the mass of the particle, and Plank’s constant. First we will convert the units of Planck's constant and uncertainty in position then by substituting all values in the Heisenberg uncertainty formula we can determine the uncertainty in velocity.

Complete solution:
The Heisenberg uncertainty formula is given as follows:
δxΔv= h4\pim{\delta x}{\Delta v = }\,\dfrac{{\text{h}}}{{{\text{4\pi m}}}}
δx{\delta x} is the uncertainty in the position of the particle.
h is the Plank’s constant.
m is the mass of the particle.
δv{\delta v} is the uncertainty in the velocity of the particle.
First, we will convert the unit of x from Ao{{\text{A}}^{\text{o}}} to meter as follows:
1 Ao{\text{1}}\,{{\text{A}}^{\text{o}}}= 10−10 m{10^{ - 10}}\,{\text{m}}
0.1 Ao{\text{0}}{\text{.1}}\,{{\text{A}}^{\text{o}}}= 10−11 m{10^{ - 11}}\,{\text{m}}
Now, we will convert the unit of h from J to kgm2s−2{\text{kg}}{{\text{m}}^{\text{2}}}{{\text{s}}^{ - {\text{2}}}} as follows:
11 J = 1 kgm2s−2{\text{1}}\,{\text{kg}}{{\text{m}}^{\text{2}}}{{\text{s}}^{ - {\text{2}}}}
6.626× 10−346.626 \times \,{10^{ - 34}} Js = 6.626× 10−34 kgm2s−16.626 \times \,{10^{ - 34}}\,{\text{kg}}{{\text{m}}^{\text{2}}}{{\text{s}}^{ - 1}}
On substituting 9.1 ×10−319.1\, \times {10^{ - 31}} kg for mass of electron, 10−11{10^{ - 11}} m for uncertainty in position, and 6.626× 10−34 kgm2s−16.626 \times \,{10^{ - 34}}\,{\text{kg}}{{\text{m}}^{\text{2}}}{{\text{s}}^{ - 1}}for h,
δx.Δv= h4\pim{\delta x}{.\Delta v = }\,\dfrac{{\text{h}}}{{{\text{4\pi m}}}}
10−11 m ×  δv= 6.626× 10−34 kgm2s−14×3.14×9.11 ×10−31 kg{\text{1}}{{\text{0}}^{ - 11}}\,{\text{m}}\, \times \,\,{\delta v = }\,\dfrac{{6.626 \times \,{{10}^{ - 34}}\,{\text{kg}}{{\text{m}}^{\text{2}}}{{\text{s}}^{ - 1}}}}{{{\text{4}} \times 3.14 \times 9.11\, \times {{10}^{ - 31}}\,{\text{kg}}}}
  δv= 6.626× 10−34 kgm2s−110−11 m × 4×3.14×9.11 ×10−31 kg\,\,{\delta v = }\,\dfrac{{6.626 \times \,{{10}^{ - 34}}\,{\text{kg}}{{\text{m}}^{\text{2}}}{{\text{s}}^{ - 1}}}}{{{\text{1}}{{\text{0}}^{ - 11}}\,{\text{m}}\, \times \,{\text{4}} \times 3.14 \times 9.11\, \times {{10}^{ - 31}}\,{\text{kg}}}}
 δv= 6.626× 10−34 ms−11.144 ×10−40 \,{\delta v = }\,\dfrac{{6.626 \times \,{{10}^{ - 34}}\,{\text{m}}{{\text{s}}^{ - 1}}}}{{1.144\, \times {{10}^{ - 40}}\,}}
 δv= 6.626× 10−34 ms−114.42 ×10−42 \,{\delta v = }\,\dfrac{{6.626 \times \,{{10}^{ - 34}}\,{\text{m}}{{\text{s}}^{ - 1}}}}{{14.42\, \times {{10}^{ - 42}}\,}}
 δv= 5.79× 106 ms−1\,{\delta v = }\,5.79 \times \,{10^6}\,{\text{m}}{{\text{s}}^{ - 1}}
So, the uncertainty in velocity is 5.79× 106 ms−15.79 \times \,{10^6}\,{\text{m}}{{\text{s}}^{ - 1}}.
Therefore, option (C) 5.79× 106 ms−15.79 \times \,{10^6}\,{\text{m}}{{\text{s}}^{ - 1}} is correct.

Note: According to the Heisenberg uncertainty principle it is impossible to simultaneously measurement of uncertainty of position and momentum. Here, uncertainty of velocity is very high 5.79× 106 ms−15.79 \times \,{10^6}\,{\text{m}}{{\text{s}}^{ - 1}} whereas the uncertainty in position is very small 10−11 m{\text{1}}{{\text{0}}^{ - 11}}\,{\text{m}} so, we can determine the position of this particle but cannot determine the velocity precisely. Unit conversion is important. On joule is equal to 1 Kg. meter square/ seconds square.
The product of mass of the particle and velocity is known as momentum. So, we can substitute p for v/m. We can use the Heisenberg uncertainty formula to determine the momentum of the particle also.
δx.Δp= h4\pi{\delta x}{.\Delta p = }\,\dfrac{{\text{h}}}{{{\text{4\pi }}}} ; Where, p{\text{p}}is the momentum.