Atomic Structure Question 190
Question: 1In 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 x $ {10^{-34}} $ kg $ m^{2} $ /s, mass of electron, $ e_{m} $ = 9.1 x $ 10^{31} $ kg)
Options:
A) $ 1.52\times {10^{-4}}m $
B) $ 5.10\times {10^{-3}}m $
C) $ 1.92\times {10^{-3}}m $
D) $ 3.84\times {10^{-3}}m $
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Answer:
Correct Answer: C
Solution:
- By Heisenberg’s uncertainty principle $ \Delta x(m\Delta V)=\frac{h}{4\pi } $
$ \Delta V_{acc}=0.005% $ Or $ \Delta V=\frac{600\times 0.005}{100}=0.03m{s^{-1}} $
$ \Delta x\times 9.1\times {10^{-31}}\times 0.03=\frac{6.6\times {10^{-34}}}{4\times 3.14} $
Hence, $ \Delta x=\frac{6.6\times {10^{-34}}}{4\times 3.14\times 0.03\times 9.1\times {10^{-31}}}=1.92\times {10^{-3}}m $
