Question

how large can the kinetic energy of an electron be that is localized within a distance (change in) x = .1 nmapproximately the diameter of a hydrogen atom (ev)

Answer 1

Answer:

The kinetic energy of an electron is $1.54\times10^{-15}\ J$

Explanation:

Given that,

Distance = 0.1 nm

We need to calculate the momentum

Using uncertainty principle

$\Delta x\Delta p\geq\dfrac{h}{4\pi}$

$\Delta p\geq\dfrac{h}{\Delta x\times 4\pi}$

Where, $\Delta p$ = change in momentum

$\Delta x$ = change in position

Put the value into the formula

$\Delta p=\dfrac{6.6\times10^{-34}}{4\pi\times10^{-10}}$

$\Delta p=5.3\times10^{-23}$

We need to calculate the kinetic energy for an electron

$K.E=\dfrac{p^2}{2m}$

Where, P = momentum

m = mass of electron

Put the value into the formula

$K.E=\dfrac{(5.3\times10^{-23})^2}{2\times9.1\times10^{-31}}$

$K.E=1.54\times10^{-15}\ J$

Hence, The kinetic energy of an electron is $1.54\times10^{-15}\ J$