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//4pi_(m)xx0.05 nm`, is there any problem in defining this value.

Explanation

`Delta x=0.002 nm = 0.002xx10^(-9)m=2.0xx10^(-12)m`
`Delta x, Deltap=(h)/(4 pi) or Delta(p)=(h)/(4pi Delta x)`
`Deltap=((6.626xx10^(-34)"kg" m^(2) s^(-1)))/(4xx3.142xx(2xx10^(-12)m))=2.638xx10^(-23) kg m s^(-1)`
Actual momentum (p)`=(h)/(4 pi xx 0.05nm)=((6.626xx10^(-34)"kg"m^(2)s^(-1)))/(4xx3.142xx(5xx10^(-11)m))=1.055xx10^(-24) "kg" m s^(-1)`
Since actual momentum is smaller than the uncertainty in measuring momentum, therefore, the momentum of electron can not be defined.