Question

Show that the number of different electron states possible for a given value of \(n\) is \(2 n^{2}\)

Short answer

Answer: The number of different electron states possible for a given value of the principal quantum number (n) is 2n².

Step-by-step solution

Understanding quantum numbers

There are four quantum numbers that describe an electron in an atom: 1. Principal quantum number (n): Represents the energy level and size of an electron's orbital (n = 1, 2, 3, ...) 2. Azimuthal quantum number (l): Represents the shape of an electron's orbital (l = 0, 1, 2, ...n-1) 3. Magnetic quantum number (m): Represents the orientation of an electron's orbital in space (m = -l, -(l-1), ...0, ...(l-1), l) 4. Electron spin quantum number (m_s): Represents the electron's intrinsic angular momentum or "spin" (m_s = -1/2, 1/2)

Determining possible values for l

For a given value of n, the azimuthal quantum number (l) can have values ranging from 0 to n-1. So there are n possible values for l.

Determining possible values for m and m_s

For each value of l, the magnetic quantum number (m) can take 2l+1 values, ranging from -l to l. The electron spin quantum number (m_s) can take two values, -1/2 and 1/2. So for each value of l, there are 2(2l+1) possible electron states, considering both m and m_s values.

Calculating the total number of electron states

We will now sum the number of electron states for all possible values of l (from 0 to n-1) by calculating the following sum: \(\sum_{l=0}^{n-1} 2(2l + 1)\)

Simplifying the sum

Let's start by pulling the constant term (2) out of the sum: \(2 \sum_{l=0}^{n-1} (2l + 1)\) Now let's split the sum into two separate sums: \(2( \sum_{l=0}^{n-1} 2l + \sum_{l=0}^{n-1} 1)\) The first sum is an arithmetic series, so we can simplify it using the formula: \(\sum_{l=0}^{n-1} a_l = n\big(\frac{a_0 + a_{n-1}}{2}\big)\). In this case, \(a_0 = 0\) and \(a_{n-1} = 2(n-1)\), so the first sum simplifies to: \(n\big(\frac{0 + 2(n-1)}{2}\big) = n(n-1)\) The second sum is simply adding 1 n times, so it simplifies to n. So, the sum can be written as: \(2(n(n-1) + n)\)

Final simplification

Now, we simplify the expression inside the parentheses and distribute the 2: \(2(n^2 - n + n) = 2n^2\) And that's it! We have shown that the number of different electron states possible for a given value of n is indeed \(2 n^2\).