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

Question

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.

Answer 1

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.

Answer 2

The comments have told you the answer, but since nobody wrote it out as an answer yet, here it is.

1. Wave motion in general has both phase velocity and group velocity. Phase velocity is the velocity at which wavefronts of a single-frequency wave move. It is equal to $v_p = f \lambda$ where $f$ is the frequency and $\lambda$ is the wavelength. Group velocity is the velocity at which the changes in wave amplitude move along; this is also the velocity of the peak of a wavepacket, and it is the velocity at which energy and momentum (and other information) can be transferred from one place to another by the waves. It is equal to $v_g = d\omega/dk$ where $\omega = 2 \pi f$ and $k = 2\pi/\lambda$.

2. The fundamental and most central statement about de Broglie waves is the way the wavelength relates to the momentum. This is $$ \lambda = \frac{h}{p} $$ When in doubt, you should make this your starting-point. Particles correspond roughly to wavepackets, so they move at the group velocity of these waves. So the quick route to your answer is to take $v = p/m$ and thus note that when $v$ increases, so does $p$, and therefore $\lambda$ decreases.

The above equation can also be written $$ p = \hbar k $$ and this is how professional physicists mostly choose to write it.

3. If you want to see some more detail, then you can bring in $E = h f = \hbar \omega$. Treating this in the limit of low speeds (where we don't need relativity), you must use the relationship $E = p^2/2m$, which gives $$ \hbar \omega = \frac{ \hbar^2 k^2 }{2m} . $$ Differentiating this, we find the group velocity is $$ v_g = \frac{d\omega}{dk} = \frac{\hbar k}{m} = \frac{p}{m} = v $$ where $v$ is the particle speed.

4. For more generality, let the energy $E$ be the total relativistic energy. In this case we have $$ E^2 - p^2 c^2 = m^2 c^4 $$ which gives $$ \omega^2 - k^2 c^2 = m^2 c^4 / \hbar^2 $$ and therefore $$ 2\omega \frac{d\omega}{dk} = 2 k c^2$$ so $$ v_g = \frac{d \omega}{dk} = \frac{k c^2}{\omega} = \frac{p c^2}{E} $$ where, don't forget, $p$ is the momentum and $E$ is the total relativistic energy. Using standard relativity, the combination $p c^2/E$ for a particle is equal to the speed of the particle.

5. Finally, then, what is the phase velocity of de Broglie waves? It is $\omega/k = E/p$. This can be a very large speed---it can even be greater than the speed of light! Also, its value depends on how you choose the zero of energy. If you are disconcerted by this, the main lesson is that the phase velocity of de Broglie waves is not directly relevant to any physical phenomenon. (It can assigned a somewhat abstract interpretation in a relativistic treatment, but I won't get into that.)

Answer 3

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