Chapter 5: Q80CE (page 193)

Determine the particle’s most probable position.

Short Answer

Expert verified

The most probable position for the particle with the given wave function is $\frac{3}{2 a}$ .

Step by step solution

Define the most probable position

A particle of mass m is represented by a wave function,

$ψ (x) = {\begin{cases} 2 \sqrt{a^{3}} x e^{- a x} & x > 0 \\ 0 & x < 0 \end{cases}$

The most probable position of this particle is mathematically defined as

$x' = \int_{- \infty}^{\infty} ψ^{2} (x) x d x$

Determine the most probable position for the given wave function.

Expanding the integral into two parts,

$\begin{array}{rcl} x' & = & \int_{- \infty}^{0} ψ^{2} (x) x d x + \int_{0}^{\infty} ψ^{2} (x) x d x \\ = & 0 + \int_{0}^{\infty} {[2 \sqrt{a^{3}} x e^{- a x}]}^{2} x d x \\ = & \int_{0}^{\infty} [4 a^{3} x^{2} e^{- 2 a x}] x d x \\ = & 4 a^{3} \int_{0}^{\infty} [x^{3} e^{- 2 a x}] d x \end{array}$

Taking the integration part separately and solving it using By Parts method

$\begin{array}{rcl} I & = & \int_{0}^{\infty} [x^{3} e^{- 2 a x}] d x \\ = & {\frac{x^{3} e^{- 2 a x}}{- 2 a} - \frac{3 x^{2} e^{- 2 a x}}{4 a^{2}} + \frac{3 x e^{- 2 a x}}{4 a^{3}} - \frac{3 e^{- 2 a x}}{8 a^{4}}|}_{0}^{\infty} \\ = & 0 - (- \frac{3 e^{0}}{8 a^{4}}) \\ = & \frac{3}{8 a^{4}} \end{array}$

Inserting this result in the most probable expression,

$\begin{array}{rcl} x' & = & 4 a^{3} (\frac{3}{8 a^{4}}) \\ = & \frac{3}{2 a} \end{array}$

Hence the most probable position for the particle with a given wave function is $\frac{3}{2 a}$ .

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Determine the particle’s most probable position.

Short Answer

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Define the most probable position

Determine the most probable position for the given wave function.

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