Question

In Young's double slit experiment, an electron beam is used to produce interference fringes of width \( \beta_1 \). Now the electron beam is replaced by a beam of protons with the same experimental setup and same speed. The fringe width obtained is \( \beta_2 \). The correct relation between \( \beta_1 \) and \( \beta_2 \) is

• A. \( \beta_1 = \beta_2 \)
• B. No fringes are formed
• C. \( \beta_1 < \beta_2 \)
• D. \( \beta_1 > \beta_2 \)

Answer 1

The Correct Option is D

The fringe width \( \beta \) in a Young's double-slit experiment is given by the formula:
\( \beta = \frac{\lambda D}{d} \) where
\( \lambda \) is the wavelength of the particles (electrons or protons), 
\( D \) is the distance between the slits and the screen, and 
\( d \) is the slit separation. 

Since the wavelength \( \lambda \) of a particle depends on its momentum, the electron, being lighter, has a longer de Broglie wavelength than the proton. Therefore, the fringe width for the electron beam \( \beta_1 \) will be greater than that for the proton beam \( \beta_2 \), 
i.e., \( \beta_1 > \beta_2 \).