Electrons emerge from an electron gun with a speed of 2.0 x 10^6 m/s and then pass through a pair of thin parallel slits. Interference fringes with...

Saturday, February 13, 2010

Electrons emerge from an electron gun with a speed of 2.0 x 10^6 m/s and then pass through a pair of thin parallel slits. Interference fringes with...

Electrons emerge from an electron gun with a speed of 2.0 x 10^6 m/s and then pass through a pair of thin parallel slits. Interference fringes with a spacing of 2.7 mm are detected on a screen far from the double slit and fairly close to the center of the pattern. The mass of electrons is 9.11 x 10^-31 kg and the mass of neutrons is 1.67 x 10^-27 kg. The fringe spacing has to be determined if the electrons are replaced by neutrons with the same speed.

The de Broglie wavelength of a particle with mass m is $\displaystyle{L}=\frac{{h}}{{p}},$ where h is the Planck's associated and equal to $\displaystyle{6.6}\cdot{{10}}^{{-{{34}}}}$ J*s and p is the momentum of the particle.

The fringe width $\displaystyle\partial{X}$ is given by $\displaystyle\partial{X}={L}\cdot\frac{{D}}{{d}},$ where D is the distance from the screen and d is the slit separation.

$\displaystyle\partial{X}={\left(\frac{{h}}{{p}}\right)}\cdot{\left(\frac{{D}}{{d}}\right)}$

For electrons, $\displaystyle\partial{X}_{{e}}={\left(\frac{{h}}{{{m}_{{e}}\cdot{v}_{{e}}}}\right)}\cdot{\left(\frac{{D}}{{d}}\right)}$

For protons, $\displaystyle\partial{X}_{{p}}={\left(\frac{{h}}{{{m}_{{p}}\cdot{v}_{{p}}}}\right)}\cdot{\left(\frac{{D}}{{d}}\right)}$

The ratio of the two gives $\displaystyle\frac{{\partial{X}_{{e}}}}{{\partial{X}_{{p}}}}=\frac{{{\left(\frac{{h}}{{{m}_{{e}}\cdot{v}_{{e}}}}\cdot{\left(\frac{{D}}{{d}}\right)}\right)}}}{{{\left(\frac{{h}}{{{m}_{{p}}\cdot{v}_{{p}}}}\cdot{\left(\frac{{D}}{{d}}\right)}\right)}}}$

=> $\displaystyle\frac{{\partial{X}_{{e}}}}{{\partial{X}_{{p}}}}=\frac{{{m}_{{p}}\cdot{v}_{{p}}}}{{{m}_{{e}}\cdot{v}_{{e}}}}$

The value of $\displaystyle\partial{X}_{{e}}={2.7}{m}{m},$ $\displaystyle{m}_{{e}}={9.11}\cdot{{10}}^{{-{{31}}}}$ kg, $\displaystyle{m}_{{p}}={1.67}\cdot{{10}}^{{-{{27}}}}$ kg

As the velocity of the electron and the proton emerging from the gun is the same $\displaystyle{v}_{{e}}={v}_{{p}},'{t}{h}{i}{s}{g{{i}}}{v}{e}{s}$ delX_p = delX_e*(m_e/m_p)

$\displaystyle\partial{X}_{{p}}={\left({2.7}\cdot{{10}}^{{-{{3}}}}\right)}{\left(\frac{{{9.11}\cdot{{10}}^{{-{{31}}}}}}{{{1.67}\cdot{{10}}^{{-{{27}}}}}}\right)}$