Question

The uncertainty in position of a proton confined to the nucleus of an atom is roughly the diameter of the nucleus. If this diameter is 7.8x10^-15 m, what is the uncertainty in the proton's momentum? Express your answer using two significant figures.

Answer 1

Answer: The uncertainty in proton's momentum is
`1.3* 10^(-20)kg.m/s`

Step-by-step explanation:

The equation representing Heisenberg's uncertainty principle follows:

`\Delta x.\Delta p\geq (h)/(2\pi)`

where,

`\Delta x` = uncertainty in position = d =
`7.8* 10^(-15)m`

`\Delta p` = uncertainty in momentum = ?

h = Planck's constant =
`6.627* 10^(-34)kgm^2/s^2`

Putting values in above equation, we get:

`\Delta p=(6.627* 10^(-34)kgm^2/s^2)/(2* 3.14* 7.8* 10^(-15)m)\\\\\Delta p=1.3* 10^(-20)kg.m/s`

Hence, the uncertainty in proton's momentum is
`1.3* 10^(-20)kg.m/s`