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

Answer:

$\Delta x = \frac{\lambda}{4\pi}$

Explanation:

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

$P = \frac{h}{\lambda}$

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

$\Delta x \times \Delta P = \frac{h}{4\pi}$

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

$\Delta x = \frac{h}{4\pi \Delta P}$

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

$\Delta x = \frac{h}{4\pi (\frac{h}{\lambda})}$

$\Delta x = \frac{\lambda}{4\pi}$

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