Question

A free particle has kinetic energy equal to 35eV a)- What is the velocity of the particle? b)- If this velocity is known to within 0.2% accuracy, what is r of the position that particle? Assume that the mass of the particle is 2 x 10^-26 kg. Use h and give the answer in nm.

Answer 1

Given:

kinetic energy of free particle, KE = 35ev

1eV =
`1.6* 10^(-19)` J

mass of the particle, m =
`2* 10^(-26)` Kg

accuracy in velocity= 0.2%= 0.002

Solution:

a) We know that

KE =
`(1)/(2)mv^(2)`

v =
`\sqrt{(2KE)/(m)}`

⇒ v =
`\sqrt{(2* 35*1.6* 10^(-19) )/(2* 10^(-26))}`

v =
`2.36*10^(4)` m/s

b) From Heisenberg's uncertainity principle:

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

`\Delta x.(mv) = (h)/(4\pi)`

`\Delta x = (h)/(4\pi* 2* 10^(-26)*2.36 * 10^(4)* 0.002)`

`\Delta x = 0.56nm`