Question

Question: Problem 42 in Chapter 9 described an optical atomic trap. In one experiment, a gas of500sodium atoms is confined in a volume of$\mathbf{1000}\mathit{\mu }{\mathbf{\text{m}}}^{\mathbf{3}}$ . The temperature of the system is500. Compute the probability that, by chance, these 500 slowly moving sodium atoms will all congregate in the left half of the available volume. Express your answer in scientific notation.

Short answer

The probability is 0

Step-by-step solution

Given data

A gas of 500 sodium atoms is confined in a volume of $1000\mu {\text{m}}^3$. The temperature of the system is 500.

Concept used

The probability of having first 500 atoms in the left half of the optical trap is $\frac{1}{2}$.

Assume that sodium behaves ideally, and the chance of the presence of a second sodium atom in the left half of the optical trap is $\frac{1}{2}$.

The chance of the presence of both atoms in the left half is equal to the product of the probability of the first and second atom on the left half of the optical trap $=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$.

Calculate the probability

$p$- the probability of the presence of all 500 atoms on the left half of the optical trap

$p={\left(\frac{1}{2}\right)}^{500}$

Apply log on both sides

$\begin{array}{c}\log p=500\log \frac{1}{2}\\ =500\times (-0.30103)\\ =-150.51\end{array}$

Use antilog to get the value of $p$

$\begin{array}{c}p={10}^{-150.51}\\ ={10}^{0.49}\times {10}^{-151}\\ =3\times {10}^{-151}\\ =0\end{array}$