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Q. The uncertainty in velocity of an electron present in the nucleus of diameter $10^{-15} m$ hypothetically should be approximately

Structure of Atom

Solution:

$\Delta x=10^{-15} m$
$m=9.1 \times 10^{-31} kg$
$\Delta x \times m \Delta v =\frac{ h }{4 \pi}$
$\Rightarrow \Delta v =\frac{6.6 \times 10^{-34} kg\, m ^{2} s ^{-1}}{10^{-15} \times 9.1 \times 10^{-31} \times 4 \times 3.14}$
$=\frac{6.6}{114.296} \times 10^{12} m / s$
App. $=0.5 \times 10^{11} m / s$