Consider a particle of mass m which can move freely along the x axis from -a/2 to a/2, but which is strictly prohibited from being found outside this region. The wave function of the particle within the allowed region is

  φ = B sin (2π n/a   x)

Explanation:

In quantum mechanics when a particle moves freely it implies that the potential is zero (V = 0), so its wave function is

     φ = A cos kx + B sin kx

we must place the boundary conditions to determine the value of the constants A and B.

In our case we are told that the particle cannot be outside the boundary given by x = ± a / 2

therefore we must make the cosine part zero, for this the constant A = 0, the wave function remains

    φ = B sin kx

the wave vector is

      k = 2π /λ

now let's adjust the period, in the border fi = 0 therefore the sine function must be zero

         φ (a /2) = 0

          0 = A sin (2π/λ  a/2)

therefore the sine argument is

          2π /λ   a/2 = n π

          λ= a / n

we substitute

          φ = B sin (2π n/a   x)

3

step-by-step explanation:

10 - 2.8 = 7.2

2 2/5 = 2.4

7.2 / 2.4 = 3

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