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strojnjashka [21]
3 years ago
8

A proton is confined within an atomic nucleus of diameter 3.60 fm. part a estimate the smallest range of speeds you might find f

or a proton in the nucleus.
Physics
1 answer:
Cerrena [4.2K]3 years ago
3 0
The answer for this problem would be:
Assuming non-relativistic momentum, then you have: 
ΔxΔp = mΔxΔv = h / (4) 
Δv = h / (4πmΔx) 
m ~ 1.67e-27 h ~ 6.62e-34,Δx = 4e-15 --> 
Δv ~ 6.62e-34 / (4π * 1.67e-27 * 4e-15) ~ 7,886,270 m/s ~ 7.89e6 m/s 
That's about 1% of the speed of light, the assumption that it's non-relativistic.
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\vec{R} = \sqrt{A^{2} + 2ABcos\theta + B^{2}}

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(m +m')v_{final} = (35\times 10^{3})v_{final}

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2(mv)(m'v')cos20^{\circ} = 2(15\times 10^{3}\times 770)(20\times 10^{3}\times 1020)cos20^{\circ} = 4.43\times 10^{14}

Now, substituting the suitable values in eqn (1), we get:

v_{final} = 900.06\ m/s

Now, the direction for the two vectors is given by:

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\theta = sin^{- 1} \frac{20\times 10^{3}\times 1020sin20^{\circ}}{(35\times 10^{3})\times 900.06} = 0.2215^{\circ}

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