Question

Young’s double slit experiment is performed using a beam of $C_{60}$ (fullerene) molecules, each molecule being made up of 60 carbon atoms. When the slit separation is 50 nm, fringes are formed on a screen kept at a distance of 1 m from the slits. Now, the experiment is repeated with $C_{70}$ molecules with a slit separation of 92.5 nm. The kinetic energies of both the beams are the same. The position of the 4$^{\text{th}}$ bright fringe for $C_{60}$ will correspond to the $n^{\text{th}}$ bright fringe for $C_{70}$. What is the value of $n$ (rounded off to the nearest integer)?

• A. 5
• B. 6
• C. 7
• D. 8

Hint:

For experiments involving different molecules, the key is comparing the fringe positions using the ratio of slit separations, as the de Broglie wavelength for identical kinetic energies will be the same.

Answer 1

The Correct Option is D

The fringe separation in Young's double slit experiment is given by: \[ y = \frac{\lambda L}{d}, \] where $\lambda$ is the wavelength, $L$ is the distance to the screen, and $d$ is the slit separation.
Since the kinetic energies are the same for both beams, the velocities of the $C_{60}$ and $C_{70}$ molecules will be the same, meaning the de Broglie wavelengths for both beams are the same. Therefore, the ratio of fringe separations for the two cases will be the ratio of the slit separations: \[ \frac{y_2}{y_1} = \frac{d_2}{d_1} = \frac{92.5 \text{ nm}}{50 \text{ nm}} = 1.85. \] The position of the 4$^{th}$ bright fringe for $C_{60}$ will correspond to the $n^{th}$ bright fringe for $C_{70}$, so we have: \[ n = 4 \times 1.85 = 7.4 \approx 8. \] Thus, the correct answer is (D) 8.