Question

The minimum uncertainty in de-Broglie wavelength of an electron accelerated from rest by a
potential difference of 6.0V, if uncertainty in measuring the position is 1/pie nm, is

(A) 6.25 Å
(B) 6.0 Å
(C) 0.625 Å
(D) 0.3125 Å​

Answer 1

Given:

Electron accelerated from rest by a potential difference of 6.0V, uncertainty in measuring the position is 1/pie nm.

To find:

Minimum uncertainty of De-Broglie wavelength.

Calculation:

Wavelength of the De-Broglie wave is:

$\lambda = \sqrt{ \dfrac{150}{6} } = \sqrt{25} = 5 \: {A}^{ \circ}$

Now , we know that , De-Broglie wavelength is also given as:

$\therefore \: \lambda = \dfrac{h}{p}$

$= > p = \dfrac{h}{ \lambda}$

$= > | \Delta p | = \dfrac{h}{ {\lambda}^{2} } \times \Delta \lambda$

Now , considering Heisenberg's Uncertainty Principle:

$\therefore \: \Delta x \times \Delta p \geqslant \dfrac{h}{4\pi}$

$= > \: \Delta x \times ( \dfrac{h}{{\lambda}^{2} } \times \Delta \lambda) \geqslant \dfrac{h}{4\pi}$

$= > \: \dfrac{ {10}^{ - 9} }{\pi} \times \bigg \{\dfrac{h}{{(5\times{10}^{-10})}^{2} } \times \Delta \lambda \bigg \} \geqslant \dfrac{h}{4\pi}$

$= > \: \bigg \{\dfrac{ {10}^{ - 9} }{{(5\times{10}^{-10})}^{2} } \times \Delta \lambda \bigg \} \geqslant \dfrac{1}{4}$

$= > \: \Delta \lambda \geqslant \dfrac{2.5}{4} \times {10}^{ - 10}$

$= > \: \Delta \lambda \geqslant 0.625 \times {10}^{ - 10}$

$= > \: \Delta \lambda \geqslant 0.625 {A}^{ \circ}$

So, final answer is:

$\boxed{ \bf{\Delta \lambda \geqslant 0.625 {A}^{ \circ} }}$