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 $$\displaystyle \frac{7}{22}$$ nm?

A
6˚A
B
60˚A
C
4˚A
D
40˚A

Solution

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

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