By Sunil Bhardwaj

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Size of the nucleus is of the order of $${ 10 }^{ -14 }$$m. If the electron is inside the nucleus, uncertainty in its position$$(\Delta x)$$should be $${ 10 }^{ -14 }$$m. $$\therefore Uncertainty \ in \ momentum (\Delta p) = \frac { h }{ \Delta x }$$ $$= \frac { 1.05 \times { 10 }^{ -34 }J.sec }{ { 10 }^{ -34 } }$$ $$= 1.1 \times { 10 }^{ -20 }Kg/m.sec$$ The momentum p must be comparable to this $$\therefore p \approx \Delta p = 1.1 \times { 10 }^{ -20 }Kg/m.sec$$ $$\therefore Energy \ E = \sqrt { { m }^{ 2 }{ c }^{ 4 } + { p }^{ 2 }{ c }^{ 2 } }$$ $$= \sqrt { { p }^{ 2 }{ c }^{ 2 } }$$ $$= pc$$ $$= 1.1 \times { 10 }^{ -20 }Kg/m.sec \times 3 \times { 10 }^{ 8 }m/sec$$ $$= 3.3 \times { 10 }^{ -12 }J$$ Converting this into volts, $$E = \frac { 3.3 \times { 10 }^{ -12 } }{ 1.6 \times { 10 }^{ -19 } } = 2.06 \times { 10 }^{ 7 }eV$$ This energy is very high and experiments shows that electron with such a high energy can make the molecule highly unstable. Therefore it is not possible for an electron to remain inside the nucleus.

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