Atomic Structure Question 393
Question: The uncertainty in momentum of an electron is $ 1\times {10^{-5}}kg-m/s $ . The uncertainty in its position will be ( $ h=6.62\times {10^{-34}}kg-m^{2}/s $ ) [AFMC 1998; CBSE PMT 1999; JIPMER 2002]
Options:
A) $ 1.05\times {10^{-28}}m $
B) $ 1.05\times {10^{-26}}m $
C) $ 5.27\times {10^{-30}}m $
D) $ 5.25\times {10^{-28}}m $
Show Answer
Answer:
Correct Answer: C
Solution:
According to $ \Delta x,\times \Delta p=\frac{h}{4\pi } $
$ \Delta x=\frac{h}{\Delta p\times 4\pi }=\frac{6.62\times {10^{-34}}}{1\times {10^{-5}}\times 4\times 3.14} $
$ =5.27\times {10^{-30}}m $ .
BETA
