# What happens to the Fringe Width when the velocity of electrons passing through the slit in Young's Double Slit Experiment increases?

A beam of electron is used in Young's Double Slit Experiment. The slit width is d. Then the velocity of electron is increased. What happens to the Fringe Width?

My approach:

$$\beta = \frac{\lambda d}{d}$$

$$v = \nu \lambda$$

Since velocity $v$ increases, while frequency $\nu$ (which is dependent on the source and independent of the velocity) remains constant, $\lambda$ increases.

Hence $\beta$ increases.

My friend's approach:

$$\lambda = \frac{h}{p}$$

$$p = mv$$ Since velocity $v$ increases, $p$ increases and hence $\lambda$ decreases.

Hence $\beta$ decreases.

Both the approaches contradict each other so I was wondering which approach is correct and why? Thanks.

• Your second equation is wrong. $\nu\lambda$ is equal to the phase velocity not the group velocity. Kyle gives a nice discussion of this here – John Rennie Dec 4 '15 at 9:01
• @John Okay, thanks. So the second approach is correct, right? – Ashish Gupta Dec 4 '15 at 13:21
• Yes, your friend is correct. – John Rennie Dec 4 '15 at 13:51
• I believe that the point you diverged from good physics was your assumption that the velocity and wavelength were not linked (your statement that the frequency was independent of velocity). – Jon Custer Dec 4 '15 at 15:48

Your approach isn't correct... The mistake is that the wavelength in reality decreases. For this purpose, we could use the equation

E = h*frequency;

And it's obvious that faster electrons have a higher energy. We know that high energies are associated with low wavelengths. Therefore the wavelength of the speeding electron will decrease, amd thus the observed fringe width would decrease.

Please correct me if there's a mistake or if I'm too vague!

• Please be sure about the principle involved in the question. This question is about Young's Double-slit experiment, which does not involve the Planck–Einstein relation much, if at all. – Dlamini Mar 4 at 6:21

If we consider electron as a particle them it has de Broglie's wavelength. And if we consider electron as a wave then it has wavelength= v/frequency. But Young's experiment tells about wave nature so Fringe width should increase.

You're wrong....It should increase..... $$v=f\lambda$$ $$\lambda=v/f$$ As $v$ increases $\lambda$ increases and hence $\beta$ increases.