Question

The uncertainty in position of a particle is same as its de-Broglie wavelength, uncertainty in its momentum is \(\ldots \ldots\) (A) \((\overline{\mathrm{h}} / \lambda)\) (B) \((2 \overline{\mathrm{h}} / 3 \lambda)\) (C) \((\lambda \lambda) \overline{\mathrm{h}}\) (D) \((3 N 2 \lambda) \overline{\mathrm{h}}\)

Short answer

The uncertainty in momentum is approximately \(\Delta p \geq \frac{2\hbar}{3\lambda}\). Option (B) is the correct answer.

Step-by-step solution

Write down the Heisenberg Uncertainty Principle

The Heisenberg Uncertainty Principle states that: \[ \Delta x \Delta p \geq \frac{\hbar}{2} \] Where - \(\Delta x\) is the uncertainty in position, - \(\Delta p\) is the uncertainty in momentum, and - \(\hbar\) is the reduced Planck constant (\(\hbar = \frac{h}{2\pi}\)).

Write down the de Broglie wavelength formula

The de Broglie wavelength formula is given by: \[ \lambda = \frac{h}{p} \] Where - \(\lambda\) is the wavelength, - \(h\) is the Planck constant, and - \(p\) is the momentum.

Find the uncertainty in position

We are given that the uncertainty in position (\(\Delta x\)) is the same as the de Broglie wavelength (\(\lambda\)): \[ \Delta x = \lambda \]

Substitute the uncertainty in position into the Heisenberg Uncertainty Principle

Now, we can substitute the value of \(\Delta x\) into the Heisenberg Uncertainty Principle equation: \[ \lambda \Delta p \geq \frac{\hbar}{2} \] Next, we'll solve for the uncertainty in momentum (\(\Delta p\)).

Solve for the uncertainty in momentum

To solve for \(\Delta p\), we'll divide both sides of the inequality by \(\lambda\): \[ \Delta p \geq \frac{\hbar}{2\lambda} \] Now we have an inequality for the uncertainty in momentum.

Compare the inequality with the given options

We need to compare the inequality we obtained for the uncertainty in momentum (\(\Delta p \geq \frac{\hbar}{2\lambda}\)) with the given options: (A) \(\frac{\hbar}{\lambda}\) (B) \(\frac{2\hbar}{3\lambda}\) (C) \(\lambda\hbar\) (D) \(3N2\lambda\hbar\) The option closest to our result is (B) \(\frac{2\hbar}{3\lambda}\). Therefore, the uncertainty in momentum is approximately \(\frac{2\hbar}{3\lambda}\).