A nonrelativistic electron is confined to a length of 500 pm on the x-axis. What is the kinetic energy of the electron if its speed is equal to the...

Hello!

Again, we have a deal with the uncertainty principle for a position and a momentum. Mathematically it is expressed as

`Delta p*Delta x gt= h/(4 pi),`

where `Delta p = m*Delta v` is an uncertainty in a momentum and `Delta x` is the uncertainty in a position.

Also we know that the kinetic energy of an electron is equal to `(m_(el) v^2)/2,` and it is given that the speed is the same as the...

Hello!

Again, we have a deal with the uncertainty principle for a position and a momentum. Mathematically it is expressed as

`Delta p*Delta x gt= h/(4 pi),`

where `Delta p = m*Delta v` is an uncertainty in a momentum and `Delta x` is the uncertainty in a position.

Also we know that the kinetic energy of an electron is equal to `(m_(el) v^2)/2,` and it is given that the speed is the same as the minimum uncertainty of a speed. So the kinetic energy is equal to

`E_k = (m_(el) v^2)/2 =(m_(el) (Delta v)^2)/2 = m_(el)/2 * (h/(4 pi) * 1/(m_(el)*Delta x))^2 =1/2 (h/(4 pi))^2 1/(m_(el)*(Delta x)^2).`

All values are given, and we can compute the result in Joules (note that p means pico- means `10^(-12)` ):

`E_k = 1/2*(1.055*10^(-34))^2/(9.11*10^(-31)*25*10^(-20)) = (1.055)^2/(2*9.11*25)*10^(-17) approx 0.00244*10^(-17) (J).`

To convert this to `eV` we have to divide by `eV` value:

`E_k =0.00244*10^(-17) / (1.60*10^(-19)) = 0.00244*10^2/1.60 approx0.153 (eV).`

So your hypothesis is correct, the answer is about `0.15 eV,` E.