Atomic Structure Question 390
Question: The position of both, an electron and a helium atom is known within 1.0 nm. Further the momentum of the electron is known within $ 5.0\times {10^{-26}}kg,m{s^{-1}} $ . The minimum uncertainty in the measurement of the momentum of the helium atom is
Options:
A) $ 50kgm{s^{-1}} $
B) $ 80kgm{s^{-1}} $
C) $ 8.0\times {10^{-26}}kgm{s^{-1}} $
D) $ 5.0\times {10^{-26}}kgm{s^{-1}} $
Show Answer
Answer:
Correct Answer: D
Solution:
- According to Heisenberg uncertainty principle, $ \Delta x\times \Delta p=\frac{h}{4\pi } $ (which is constant). As $ \Delta x $ for electron and helium atom is same, thus momentum of electron and helium will also be same, therefore the momentum of helium atom is equal to $ 5\times {10^{-26}}kgm{s^{-~1}}. $
