Question

In Young's double slit experiment, an electron beam is used to produce interference fringes of width β1 . Now the electron beam is replaced by a beam of protons with the same experimental set-up and same speed. The fringe width obtained is β2 . The correct relation between β1 and β2 is

• A. β₁=β₂
• B. No fringes are formed
• C. β₁<β₂
• D. β₁>β₂

Answer 1

The Correct Option is D

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:

The fringe width \( \beta \) in Young's double-slit experiment is given by the formula:

\[ \beta = \frac{\lambda D}{d} \] where:

• \( \lambda \) is the wavelength of the particle,
• \( D \) is the distance between the screen and the slits,
• \( d \) is the distance between the slits.

 

The de Broglie wavelength \( \lambda \) of a particle is given by:

\[ \lambda = \frac{h}{mv} \] where:

• \( h \) is Planck's constant,
• \( m \) is the mass of the particle,
• \( v \) is the velocity of the particle.

 

Since protons are much heavier than electrons, their de Broglie wavelength will be smaller. This results in a smaller fringe width for protons compared to electrons. Hence, we expect:

\[ \beta_1 > \beta_2 \]

Correct Answer: (D) \( \beta_1 > \beta_2 \) 

Answer 2

In Young's double slit experiment, the fringe width \( \beta \) is given by the formula: \[ \beta = \frac{\lambda D}{d} \] where: \( \lambda \) = wavelength of the particle (electron or proton), \( D \) = distance between the slits and the screen, \( d \) = distance between the two slits. The fringe width depends on the wavelength of the particles involved. Since electrons and protons are both charged particles, they exhibit wave-like behavior, but their wavelengths are different. The wavelength \( \lambda \) is inversely proportional to the momentum of the particle, which can be given by the de Broglie relation: \[ \lambda = \frac{h}{p} \] where \( h \) is Planck's constant and \( p \) is the momentum of the particle. For the same speed (momentum), protons have much greater mass compared to electrons. Thus, the de Broglie wavelength of protons is much smaller than that of electrons. Since the wavelength \( \lambda \) for protons is smaller than that for electrons, the fringe width \( \beta_2 \) for protons will be smaller than the fringe width \( \beta_1 \) for electrons. Therefore, the correct relation between \( \beta_1 \) and \( \beta_2 \) is: \[ \beta_1 > \beta_2 \] Thus, the fringe width obtained for protons will be smaller than that obtained for electrons.