Question

If an electron has a wavelength $\lambda$ and it has an uncertainty in its kinetic energy equal to $\Delta E$, then what will be uncertainty in its position?

• A. $\frac{h^2}{4\pi m\lambda \Delta E}$
• B. $\frac{h}{4\pi m\lambda \Delta E}$
• C. $\frac{h}{4\pi \lambda \Delta E}$
• D. $\frac{h^2}{4\pi \lambda \Delta E}$

Answer 1

The correct option is A $\frac{h^2}{4\pi m\lambda \Delta E}$

According to Heisenberg uncertainity principle,
$\Delta x\times \Delta v=\frac{h}{4\pi m}$
As $KE=\frac{1}{2}m{v}^2$
For uncertainity in K.E.
$\delta KE=mv\delta v$
$\Rightarrow \delta KE=\frac{h}{\lambda }\delta v$ $(Given\Delta K.E.=\Delta E)$
$\Rightarrow \delta v=\frac{\lambda }{h}\Delta E$
So, $\Delta x\times \Delta v=\Delta x\times \frac{\lambda }{h}\Delta E=\frac{h}{4\pi m}$
$\Rightarrow \Delta x=\frac{h^2}{4\pi m\lambda \Delta E}$