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

The kinetic energy of an electron is
`1.54*10^(-15)\ J`

Step-by-step explanation:

Given that,

Distance = 0.1 nm

We need to calculate the momentum

Using uncertainty principle

`\Delta x\Delta p\geq(h)/(4\pi)`

`\Delta p\geq(h)/(\Delta x* 4\pi)`

Where,
`\Delta p` = change in momentum

`\Delta x` = change in position

Put the value into the formula

`\Delta p=(6.6*10^(-34))/(4\pi*10^(-10))`

`\Delta p=5.3*10^(-23)`

We need to calculate the kinetic energy for an electron

`K.E=(p^2)/(2m)`

Where, P = momentum

m = mass of electron

Put the value into the formula

`K.E=((5.3*10^(-23))^2)/(2*9.1*10^(-31))`

`K.E=1.54*10^(-15)\ J`

Hence, The kinetic energy of an electron is
`1.54*10^(-15)\ J`