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.'

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now