Question

Using the Heisenberg uncertainty principle arrange the following particles in the order of increasing lowest energy possible.
$\left(I\right)$ an electron in $H_2$ molecule
$\left(II\right)$ a $H$ atom in a $H_2$ molecule
$\left(III\right)$ a proton in the carbon nucleus
$\left(IV\right)$ a $H_2$ molecule within a nanotube

• A. $\left(I\right)<\left(III\right)<\left(II\right)<\left(IV\right)$
• B. $\left(IV\right)<\left(II\right)<\left(I\right)<\left(III\right)$
• C. $\left(II\right)<\left(IV\right)<\left(III\right)<\left(I\right)$
• D. $\left(IV\right)<\left(I\right)<\left(II\right)<\left(II\right)$

Answer 1

Answer: (A)

$\left(I\right)<\left(III\right)<\left(II\right)<\left(IV\right)$

We know that

nanotude $\simeq nm\simeq {10}^{-6}m$

$\Delta x5$ more $\Rightarrow AP\simeq$ minimum $\Rightarrow P$ is moles

$\Rightarrow E=\frac{P^v}{2m}5$ minimum

So $\left(A\right)$ and $\left(cC\right)$ are ruled one

$H$ atom is $H_2$ molecule is order of molecular size $\Delta x_1\simeq$ molecular size

electron in $H_2$ molecule

$\Delta x_2=$ atoms size

$\Delta x_1>\Delta x_2$

$\Rightarrow E_1<E_2$

Hence option $\left(B\right)$