Looking for classes? Ksquare Career Institute, Bengaluru →
If uncertainty in position and momentum are equal, then uncertainty in velocity is -
A
$\dfrac{1}{m}\sqrt{\dfrac{h}{\pi}}$
B
$\sqrt{\dfrac{h}{\pi}}$
C
$\dfrac{1}{2m}\sqrt{\dfrac{h}{\pi}}$
D
$\sqrt{\dfrac{h}{2\pi}}$
Detailed Solution
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 $\dfrac{1}{2m}\sqrt{\dfrac{h}{\pi}}$.
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 $\dfrac{1}{2m}\sqrt{\dfrac{h}{\pi}}$.