A proton is confined within an atomic nucleus of diameter 3.80 fm (1fm=10^−15m). Estimate the smallest range of speeds you might find for a proton in the nucleus.

0m/s to 8.29×20^6m/s

Explanation:

The proton is confined in 3.80 fm

then the maximum uncertainty is

Δx=3.810^−15m

Uncertainty principle can be expressed as

Δp. Δx ⩾ h/4π eqn(1)

Δp ⩾ h/4π.Δx

p= mv which is the uncertainty in momentum

m.Δv ⩾ h/4π.Δx

We can also express it in term of velocity as

Δv= h/4πmΔx eqn(2)

m=Mass of proton= 1.673 × 10^-27 kilograms

h= Plank's constant= 6.626 × 10^-34kilograms

Δv= uncertainty in speed

If we input the values into eqn(2) we have

Δv=6.626 × 10^-34/(4π× 1.673 × 10^-27 × 3.8×10^−15)

Δv=8.29×20^6m/s

Hence, the smallest range of speeds you might find for a proton in the nucleus is 0<v<8.29×20^6m/s

its a 5.5 n

glad that

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer