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Heisenberg's Uncertainty Principle
Concepts tested here
- Uncertainty in velocity from uncertainty in momentum
- Uncertainty principle
All Questions
2008 AIPMT 2 questions
-
If uncertainty in position and momentum are equal, then uncertainty in velocity is -According to the Heisenberg uncertainty principle: $\Delta p \cdot \Delta x \geq \dfrac{h}{4\pi}$
Given $\Delta x = \Delta p$, so $(\Delta p)^2 \geq \dfrac{h}{4\pi}$
Since $\Delta p = m\,\Delta v$: $(m\,\Delta v)^2 \geq \dfrac{h}{4\pi}$
$m\,\Delta v \geq \sqrt{\dfrac{h}{4\pi}} = \dfrac{1}{2}\sqrt{\dfrac{h}{\pi}}$
$\Delta v \geq \dfrac{1}{2m}\sqrt{\dfrac{h}{\pi}}$
So the uncertainty in velocity is $\dfrac{1}{2m}\sqrt{\dfrac{h}{\pi}}$. -
The measurement of the electron position is associated with an uncertainty in momentum, which is equal to $1 \times 10^{-18}$ g cm s$^{-1}$. The uncertainty in electron velocity is :
(Mass of an electron is $9 \times 10^{-28}$ g)Momentum $p = mv$, so uncertainty in momentum $\Delta p = m\,\Delta v$
Uncertainty in momentum $(m\,\Delta v) = 1 \times 10^{-18}$ g cm s$^{-1}$
Uncertainty in velocity: $\Delta v = \dfrac{\Delta p}{m}$
$\Delta v = \dfrac{1 \times 10^{-18}}{9 \times 10^{-28}}$
$\Delta v = 1.1 \times 10^{9}$ cm s$^{-1}$
$\approx 1 \times 10^{9}$ cm s$^{-1}$