Question

If the position of the electron is measured within an accuracy of ± 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is h/4π m × 0.05 nm, is there any problem in defining this value.

Answer 1

"From Heisenberg's uncertainty principle,

$\Delta x\quad \times \quad \Delta p\quad =\quad \frac { h }{4\pi}$$\Rightarrow \quad \Delta p\quad =\quad \frac { 1 }{ \Delta x } .\frac { h }{4\pi}$

Where,

$\Delta x$= uncertainty in position of the electron

$\Delta p$ = uncertainty in momentum of the electron

$\Delta x = 0.002 nm = 2\quad \times \quad { 10 }^{ -12 } \quad m (given)$

Therefore, substituting the values in the expression of Δp.

$\Delta p\quad =\quad \frac { 1 }{ 0.002\quad nm } \quad \times \quad \frac { 6.626\quad \times \quad { 10 }^{ -34 }\quad Js }{ 4\quad \times \quad \left( 3.14 \right)}$

$=\quad \frac { 1 }{ 2\quad \times \quad { 10 }^{ -12 }\quad m } \quad \times \quad \frac { 6.626\quad \times \quad { 10 }^{ -34 }\quad Js }{ 4\quad \times \quad 3.14 }$

$=\quad 2.637\quad \times \quad { 10 }^{ -23 }\quad Js{ m }^{ -1 }$

$\Delta p\quad =\quad 2.637\quad \times \quad { 10 }^{ -23 }\quad Kgm{ s }^{ -1 }\quad \left( 1\quad J\quad =\quad 1\quad Kgm{ s }^{ -1 } \right)$

$Actual\quad Momentum\quad =\quad \frac { h }{ 4\pi \quad \times \quad 0.05\quad nm }$

$=\quad \frac { 6.626\quad \times \quad { 10 }^{ -34 } }{ 4\quad \times \quad 3.14\quad \times \quad 5\quad \times \quad { 10 }^{ -11 } }$

$Momentum\quad =\quad 1.055\quad \times \quad { 10 }^{ -24 }\quad Kg.{ m }/{ sec }$"

Answer 2

Answer:

By applying the uncertainty principle,

ΔP=

4πΔx

h

=

4×3.1416×0.002×10

−9

6.626×10

−34

=2.63×10

−23

kg m/s

The uncertainty in momentum is,

4πm×0.002×10

−9

h

=

4×π×m

h×5×10

11

.

This is much larger than the actual momentum, i.e,

4πm

h

×0.05 nm. Hence, it cannot be defined.