Question

5, a. Electron speed is measured with an accuracy rate of 0.003% having a speed of 5x103 m/s. Determine the uncertainty of the position of the electron! b. The meson charge π has a silent energy of 140 MeV and a lifetime of 26ns. Calculate the uncertainty of the meson energy rt and the level of accuracy!

Answer 1

5

a.

Momentum of Electron in this case = $m_{e}V$ , where $m_{e}$ is the mass of electron [in non-relativistic domain - 9.11 x 10^(-31) Kg] and V is the velocity of electron ( 5 x 10^3 m/s )

Momentum of electron = 9.11 x 10^(-31) Kg x 5 x 10^3 m/s = 4.56 x 10^(-27) Kgm/s

Electron speed has an uncertainity of 0.003%

So the uncertainity in momentum of electron = (0.003/100) x 4.56 x 10^(-27) Kgm/s = 1.37 x 10^(-31) Kgm/s

As per Heisenberg's uncertainity principle  $\Delta P\times \Delta x = h/4\pi$

h = 6.63 x 10^(-34) Joule-sec

Hence 1.37 x 10^(-31) Kgm/s x $\Delta x$ = 6.63 x 10^(-34) Joule-sec / 4$\pi$

$\Delta x$ = 6.63 x 10^(-34) J-sec/ [ 1.37 x 10^(-31) Kgm/s x 4 x $\pi$] = 4.84 x 10^(-3) m/ 4 x $\pi$ = 3.85 x 10^(-4) metrs

= 0.385 mm

b. Heisenberg's uncertainity principle in Energy and time format is

  $\Delta E \times \Delta t = h/4\pi$

In this case life time = $\Delta t$ = 26 x 10^(-9) s

  $\Delta E$ = [6.63 x 10^(-34) J-s/ 4$\pi$]/ [ 26 x 10^(-9) s] = 2.03 x 10^(-27) J

Meson energy = 140 x 10^6 eV = 140 x 10^6 x 1.6 x 10^(-19 ) J

Uncertainity level = $(\Delta E/E)\times 100$ = {2.03 x 10^ (-27) J / [ 140 x 10^6 x 1.6 x 10^(-19)] } x 100

= 9.06 x 10^(-15) %

Level of accuracy = 2.03 x 10^(-27) J/