Question

Assuming Heisenberg Uncertainty Principle to be true, what could be the maximum uncertainty in the de-Broglie wavelength of a moving electron accelerated by a potential difference of 6 V whose uncertainty in position is $\frac{7}{22}$ nm?

• A. $6\mathring{A}$
• B. $60\mathring{A}$
• C. $4\mathring{A}$
• D. $40\mathring{A}$

Answer 1

The correct option is D $40\mathring{A}$

$\Delta p\Delta x\ge \frac{h}{4\pi }$ (Heisenberg uncertainity principle)
$\Delta p\times \frac{7}{22}\times {10}^{-9}\ge \frac{6.6\times {10}^{-34}}{4\times \frac{22}{7}}$
$\Delta p\ge \frac{6.6\times {10}^{-25}}{4}$
So, minimum uncertainity in $\Delta p$ is $\frac{h}{4}$
Minimum uncertainity in $\Delta p$ will result in maximum uncertainity in $\Delta \lambda$
De-broglie's wavelength $\Delta \lambda =\frac{h}{\Delta p}$
$\Delta \lambda =\frac{h}{\frac{h}{4}}$
$\Delta \lambda =40\mathring{A}$