Question

Werner Heisenberg considered the limits of how precisely we can measure the properties of an electron or other microscopic particle. He determined that there is a fundamental limit to how closely we can measure both position and momentum. The more accurately we measure the momentum of a particle, the less accurately we can determine its position. The converse also true. This is summed up in what we now call the Heisenberg uncertainty principle.
The equation si `deltax.delta (mv)ge(h)/(4pi)`
The uncertainty in the position or in the momentum of a marcroscopic object like a baseball is too small to observe. However, the mass of microscopic object such as an electon is small enough for the uncertainty to be relatively large and significant.
What would be the minimum uncetaintty in de-Broglie wavelength of a moving electron accelerated by potential difference of 6 volt and whose uncetainty in position is `(7)/(22)` nm?
A. `6.25 Å`
B. `6 Å`
C. `0.625 Å`
D. `0.1325 Å`

Answer 1

Correct Answer - C
`lambda_(D.B.)=sqrt(150/6) Å=5 Å`
and `Deltax.Deltap ge h/(4pi), p=h/lambda` or `Deltap=h/lambda^(2) Deltalambda`
`rArr Deltax. h/lambda^(2)xxDeltalambda ge h/(4pi)`
`rArr 1/pixx10^(-9)/lambda^(2)xxDeltalambda ge 1/(4pi)rArr Deltalambda ge 2.5/4xx10^(-10)`
`Deltalambda ge 0.625 Å`