If the position of electron is …
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Gaurav Seth 5 years, 2 months ago
Solution
By applying the uncertainty principle,
{tex}\displaystyle \Delta P = \frac {h}{4 \pi \Delta x} = \frac {6.626 \times 10^{-34} }{4 \times 3.1416 \times 0.002 \times 10^{-9} } = 2.63 \times 10^{-23}\ kg\ m/s{/tex}
The uncertainty in momentum is,
{tex}\displaystyle \frac {h}{4 \pi m \times 0.002 \times 10^{-9} } = \frac {h \times 5 \times 10^{11} }{4 \times \pi \times m}{/tex}
This is much larger than the actual momentum, i.e,
{tex}\displaystyle \frac {h}{4 \pi m} \times 0.05\ nm{/tex}. Hence, it cannot be defined.
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