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Which of the following represents a possible wavefunction for an electron?

a) Ψ = 3x + 2
b) Ψ = e⁻²ˣ
c) Ψ = sin(x)
d) Ψ = x²

User Immorality
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1 Answer

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Final answer:

The valid wavefunctions for an electron from the given options are (b) Ψ = e⁻²ˣ and (c) Ψ = sin(x), as their squares result in real, non-negative values and are normalizable over all space.

Step-by-step explanation:

A wavefunction, represented by the Greek letter psi (Ψ), describes the state of an electron in quantum mechanics. A valid wavefunction must meet certain criteria, one of which is that the square of its magnitude (|Ψ|^2) represents the probability density of finding an electron at a given point in space. This probability density must be a real, non-negative quantity because probabilities cannot be negative.

The question asks which of the following represents a possible wavefunction for an electron, given that the square of the wavefunction is always a real quantity and proportional to the probability of finding an electron at a given point:

  • (a) Ψ = 3x + 2
  • (b) Ψ = e⁻²ˣ
  • (c) Ψ = sin(x)
  • (d) Ψ = x²

Each of the functions provided must be assessed on the basis that when squared, the result should be a real, non-negative quantity. Let's analyze each option:

  • (a) 3x + 2 can result in negative values depending on the value of x.
  • (b) e⁻²ˣ, when squared, gives a real, positive number for all values of x.
  • (c) sin(x), when squared, gives real, non-negative values in the range of 0 to 1.
  • (d) x² when squared, gives a real, positive number for all values of x.

Thus, the possible wavefunctions for an electron are options (b), (c), and (d) as their squares always result in real, non-negative quantities. Since the wavefunction must be normalizable (integral over all space of |Ψ|^2 must be finite), we can also rule out (a) and (d) as they are not normalizable over all space. Hence, the correct answer is the combinations of (b) and (c).

User Rare Pleasures
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