Question

The uncertainty in a proton's position is 0.015 nm. a) What is the minimum uncertainty Δp in its momentum? b) What is the kinetic energy of a proton whose momentum is equal to this uncertainty (Δp=p)? (Unit: meV)

Answer 1

`\Delta P_(x)=3.51 *10^(-24)` kg m/s

`E = 3.68 * 10^(-21)` J

Step-by-step explanation:

given data

uncertainty in proton position
`\Delta x` = 0.015 nm

according to Heisenberg's principle of uncertainty

`\Delta x \Delta P_(x)= ((h)/(2\pi ))/(2)`

Where h is plank constant =
`6.6260 * 10^(-34)` j-s

`\Delta P_(x)= ((h)/(2\pi ))/(2\Delta x)`

`\Delta P_(x)= ((6.6260 * 10^(-34))/(2\pi ))/(2*0.015*10^(-9))`

`\Delta P_(x)=3.51 *10^(-24)` kg m/s

b) kinetic energy of proton whose momentum

`P =\Delta p`

`E =(\Delta p^(2))/(2m)`

where m is mass is proton

`E =((3.51*10^(-24))^(2))/(2*1.67*10^(-27))`

`E = 3.68 * 10^(-21)` J