Calculate Expectation Value & SD of N1, N2 in 0.1m^3 Vacuum Chambers

[agentnerdo]

The conc.of molecules in two ultra-high-vacuum chambers of volumes V1=V2=0.1m^3 is 10^6 molecules/m3

You count the # of molecules N1 and N2, in each chamber at some moment of time.

a.find the expectation value and the SD of N1 and N2 . calc $$\sigma$$$$_{N1}$$, sketch the prob distribution of N1.


I used SD= sqt(pq) x sqt(N)

=sqt(0.5x0.5) x sqt (2,000,000)
=707.1

I assume for Expectation value i use

W= $$\underline{N!}$$(v/V) (1-v/V)$$^{N-n}$$
n!(N-n)!

I am just sure if N will be 2million, or even what my n will be...

any help be appriciated!

 

[eljose79]

Hello, thank you for your question. Let me help you with the calculation of the expectation value and standard deviation for N1 and N2.

First, we need to define the variables:

N1 = number of molecules in chamber 1
N2 = number of molecules in chamber 2
V1 = volume of chamber 1 = V2 = volume of chamber 2 = 0.1 m^3
n = number of molecules in a given state
N = total number of molecules in the chamber (N = N1 + N2)

The expectation value for N1 can be calculated as:

E[N1] = \sum_{n=0}^{N} nP(n)

where P(n) is the probability of having n molecules in chamber 1, which can be calculated using the binomial distribution:

P(n) = \frac{N!}{n!(N-n)!} \left(\frac{v}{V}\right)^n \left(1-\frac{v}{V}\right)^{N-n}

Substituting the values, we get:

E[N1] = \sum_{n=0}^{N} n \frac{N!}{n!(N-n)!} \left(\frac{10^6}{0.1}\right)^n \left(1-\frac{10^6}{0.1}\right)^{N-n}

We can simplify this equation to:

E[N1] = N \left(\frac{10^6}{0.1}\right) \left(1-\frac{10^6}{0.1}\right)^{N-1}

Similarly, the expectation value for N2 can be calculated as:

E[N2] = N \left(\frac{10^6}{0.1}\right) \left(1-\frac{10^6}{0.1}\right)^{N-1}

The standard deviation for N1 and N2 can be calculated using the formula:

\sigma_{N1} = \sqrt{\sum_{n=0}^{N} (n-E[N1])^2 P(n)}

and

\sigma_{N2} = \sqrt{\sum_{n=0}^{N} (n-E[N2])^2 P(n)}

We can also simplify these equations to:

\sigma_{N1} = \sqrt{N\left(\frac{10^