Chapter 1: Q50E (page 1)

If things really do have a dual wave-particle nature, then if the wave spreads, the probability of finding the particle should spread proportionally, independent of the degree of spreading, mass, speed, and even Planck’s constant. Imagine that a beam of particles of mass m and speedv, moving in the x direction, passes through a single slit of width w . Show that the angle $θ_{1}$ at which the first diffraction minimum would be found ( $n λ = w s i n θ_{n}$ , from physical optics) is proportional to the angle at which the particle would likely be deflected $θ ≅ \frac{∆ p_{y}}{p}$ , and that the proportionality factor is a pure number, independent of m, v, w and h . (Assume small angles: $s i n θ ≅ t a n θ ≅ θ$ ).

Expert verified

It is shown that $θ_{1} = 4 πθ \Rightarrow θ_{1} αθ$

Step by step solution

It is known that the diffraction minimum can be obtained at $θ_{1} = \frac{λ}{w}$

Since, $θ_{1}$ is a very small angle, then

$θ_{1} ≅ θ_{1}$

Here, $θ_{1}$ is the angle at which the first diffraction minimum is obtained. So,

$θ_{1} = \frac{λ}{w} θ_{1} = \frac{h}{m v w}$

Now, the angle at which the particle deflected is

$θ = ≅ \frac{∆ p_{y}}{p}$ ….. (1)

Since, it is known that

$∆ y ∆ p_{y} \approx \frac{h}{2}$

And the momentum, p = mv

So, substitute the above value into the angle equation (1), and you have

$θ \approx \frac{h}{(2 ∆ y) (m v)} \approx \frac{h}{2 w m v} \approx \frac{h}{4 πmvw}$

Now, from the above calculations, you can say that $θ_{1} = 4 πθ \Rightarrow θ_{1} αθ$ and the proportionality factor is a pure number equals $4 π$ .

Hence, it is proved that $θ_{1} α θ$ and the proportionality factor is a pure number.

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