Question

Suppose that the uncertainty in the momentum of a partícle is equal to this momentum (Ap p). How is the minimum uncertainty in the position of the particle related to its de Broglie wavelength

Answer 1

`\Delta x = (\lambda)/(4\pi)`

Step-by-step explanation:

As per de Broglie theory we know that it is given as

`P = (h)/(\lambda)`

now here we can say that by the principle of uncertainty we have

`\Delta x * \Delta P = (h)/(4\pi)`

now we can use it to find the uncertainty in position as

`\Delta x = (h)/(4\pi \Delta P)`

now plug in the value of momentum as per de Broglie theory

`\Delta x = (h)/(4\pi ((h)/(\lambda)))`

`\Delta x = (\lambda)/(4\pi)`

So above is the maximum uncertainty in position in terms of de Broglie wavelength