A) \[5.79\times {{10}^{6}}m{{s}^{-1}}\]
B) \[5.79\times {{10}^{{}}}m{{s}^{-1}}\]
C) \[5.79\times {{10}^{8}}m{{s}^{-1}}\]
D) \[5.79\times {{10}^{5}}m{{s}^{-1}}\]
Correct Answer: A
Solution :
By Heisenberg's uncertainty principle |
\[\Delta p\times \Delta x\ge \frac{h}{4\pi }\] |
or \[\Delta v\times \Delta x\ge \frac{h}{4\pi m}\] |
\[\Delta p\to \,\] uncertainty in momentum |
\[\Delta x\to \,\] uncertainty in position |
\[\Delta v\to \,\] uncertainty in velocity |
\[m\to \,\] mass of particle |
Given that, |
\[\Delta x=0.1\,{\AA}=0.1\times {{10}^{-10}}\,m\] |
\[\Delta x=0.1\,{\AA}=0.1\times {{10}^{-10}}\,m\] |
\[m=9.11\times {{10}^{-31}}\,kg\] |
\[h=Planck\,\text{constant}\,=6.626\times {{10}^{-34}}\,Js\] |
\[\pi =3.14\] |
In uncertain position \[\Delta v\times \Delta x=\frac{h}{4\pi m}\] |
\[\Delta v\times 0.1\times {{10}^{-10}}=\frac{6.626\times {{10}^{-34}}}{4\times 3.14\times 9.11\times {{10}^{-31}}}\] |
\[\Delta v=\frac{6.626\times {{10}^{-34}}}{4\times 3.14\times 9.11\times {{10}^{-31}}\times 0.1\times {{10}^{-10}}}\,m{{s}^{-1}}\] |
\[=5.785\times {{10}^{6}}\,m{{s}^{-1}}\] |
\[=5.79\times {{10}^{6}}\,m{{s}^{-1}}\] |
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