Question

Suppose that you look once every second at a system with $300$oscillators and$100$ energy quanta, to see whether your favorite oscillator happens to have all the energy (all$100$ quanta) at the instant you look. You expect that just once out ofrole="math" localid="1655710247969" $1.7\times {10}^{96}$ times you will find all of the energy concentrated on your favorite oscillator. On the average, about how many years will you have to wait? Compare this to the age of the Universe, which is thought to be aboutrole="math" localid="1655710262469" $1\times {10}^{10}$ years.role="math" localid="1655710345899" $1\mathrm{Y}\approx \mathrm{\pi }\times {10}^7\mathrm{S}$

Short answer

I have to wait$5.4\times {10}^{88}$and it is$5\times {10}^{78}$times the age of the universe.

Step-by-step solution

Identification of the given data

The given data can be listed below as-

• The oscillator of the system is N = $300$.
• The energy quanta of the system is q =$100$ .
• The energy concentrated in the oscillator is 1 out of .$1.7\times {10}^{96}$

Significance of the entropy

The entropy is referred to as the measuring amount of a particular energy that is unavailable for doing a certain amount of work.

The concept of entropy gives the number of the years.

Determination of the number of years

The equation of the number of the ways to arrange the energy quanta of a system is,

$\Omega =\frac{\left(q+N-1\right)!}{q!\left(N-1\right)}$

Here, $\Omega$is the number of ways, q is the energy quanta and N is the oscillators

Substituting the values in the above equation,

role="math" localid="1655710971347" $$\begin{gathered} \Omega =\frac{\left(100+300-1\right)!}{100!\left(300-1\right)!} \\ =1.7\times {10}^{96} \end{gathered}$$

With the help of the number of ways, the average time in years can be expressed as-

$$\begin{gathered} t=\frac{1.7\times {10}^{96}}{\left(365\mathrm{days}\right)\left({\displaystyle \frac{24\mathrm{h}}{1\mathrm{day}}}\right)\left({\displaystyle \frac{60\min }{1\mathrm{h}}}\right)\left({\displaystyle \frac{60\sec }{1\min }}\right)} \\ =5.4\times {10}^{88}\mathrm{years} \end{gathered}$$

The comparison of the average time with the age of the universe is calculated as-

$$\begin{gathered} \frac{t}{t_{univ}}=\frac{5.4\times {10}^{88}\mathrm{years}}{1\times {10}^{10}\mathrm{years}} \\ =5\times {10}^{78} \end{gathered}$$

Thus, I have to wait $5.4\times {10}^{88}$and it is $5\times {10}^{78}$times the age of the universe.