Chemistry - Atomic Structure Question with Solution | TestHub

Given:

$\lambda =\pi \mathring{A}$

$\Delta \lambda =0.05\mathring{A}$

Momentum relation:

$p=\frac{h}{\lambda }$

$\Rightarrow \frac{\Delta p}{p}=\frac{\Delta \lambda }{\lambda }$

So,

$\Delta p=p\frac{\Delta \lambda }{\lambda }$

Using Heisenberg's uncertainty principle:

$\Delta x\cdot \Delta p\ge \frac{h}{4\pi }$

Substitute $p=\frac{h}{\lambda }$:

$\Delta x\ge \frac{h}{4\pi }\cdot \frac{\lambda }{h}.\frac{\lambda }{\Delta \lambda }$

$\Delta x\ge \frac{{\lambda }^2}{4\pi \Delta \lambda }$

Now substitute values:

${\lambda }^2=\pi \mathring{A}$, $\Delta \lambda =0.05\mathring{A}$

$\Delta x=\frac{\pi }{4\pi (0.05\mathring{A})}=\frac{1}{0.2}=5\mathring{A}$