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Q. The uncertainty in the position of an electron moving with a velocity of $3 \times 10^{4} cm / sec$ accurate up to $0.011 \%$ will be:

Structure of Atom

Solution:

According to Heisenberg principle

$\Delta x=\frac{h}{4 \pi m \Delta v}$

$\Delta x=\frac{6.6 \times 10^{-27} \times 100}{4 \times 3.14 \times 9.1 \times 10^{-28} \times 3 \times 10^{4} \times 0.011}$

$\Delta x=0.175\, cm$