A proton (rest mass is 1.673 times 10^-27 kg, rest energy is 938.3 MeV) has kine

Question

A proton (rest mass is 1.673 times 10^-27 kg, rest energy is 938.3 MeV) has kinetic energy of 2500 MeV. find its momentum (in kg-m/s) (use relativistic relations) find its wavelength if one measures the proton's position to an uncertainty delta x of + l-5.60 times 10^-14 m. find the minimum possible uncertainty in the proton's momentum A panicle of mass 6.646 times 10^-27 kg is confined to a one-dimensional box of length 3.0 times 10^-14 m. What wavelength photon is needed to excite this panicle from the quantum state n = 2 to the state n = 4?

Explanation / Answer

P^2 / 2m = 2500 MeV

P^2 / 2m = 2500 *(1.6*10^-19)

P^2   = 2*(1.673*10^-27)*2500 *(1.6*10^-19)

P^2   = 1.34*10^-40

P = 1.16*10^-21 kg. m/s

Wavelength = h/P

Wavelength = (6.64*10^-34) / (1.16*10^-21)

Wavelength = 5.74*10^-13 m

Delta x * Delta P >= h / 4Pi

Delta x * Delta P >= h / 4Pi

Delta P >= h / 4Pi * (Delta x)

Delta P >= [ (6.64*10^-34) / 4Pi ] * [1/ (5.6*10^-14)]

Delta P >= 9.44*10^-22 kg m/s

Energy in Tansition = 3h^2 / 8 ma^2

Energy in Tansition = 3*(6.64*10^-34)^2 / 8 (6.64*10^-27)*(3.8*10^-14)^2

Energy in Tansition = 1.72*10^-14

This equals to energy of photon

Energy in Tansition = h*c / lambda

lambda = h*c / Energy in Tansition

lambda = (6.64*10^-34)*(3*10^8) / (1.72*10^-14)

lambda = 1.15*10^-11 m

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