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. What would the fringe spacing be if the electrons were replaced by neutrons with the same speed? (mass e = 9.11 x 10^-31 kg, mass neutron = 1.67 x 10^-27 kg)

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 L = h/p, where h is the Planck's associated and equal to 6.6*10^-34 J*s and p is the momentum of the particle.
The fringe width delX is given by delX = L*D/d, where D is the distance from the screen and d is the slit separation.
delX = (h/p)*(D/d)
For electrons, delX_e = (h/(m_e*v_e))*(D/d)
For protons, delX_p = (h/(m_p*v_p))*(D/d)
The ratio of the two gives (delX_e)/(delX_p) = ((h/(m_e*v_e)*(D/d)))/( (h/(m_p*v_p)*(D/d)))
=> (delX_e)/(delX_p) = (m_p*v_p)/(m_e*v_e)
The value of delX_e = 2.7 mm, m_e = 9.11*10^-31 kg, m_p = 1.67*10^-27 kg
As the velocity of the electron and the proton emerging from the gun is the same v_e = v_p,' this gives delX_p = delX_e*(m_e/m_p)
delX_p = (2.7*10^-3)((9.11*10^-31)/(1.67*10^-27))
delX_p = 1.47*10^-6
= 1.47 mum
The fringe spacing when the neutrons are replaced by electrons is equal to '1.47 mum.'