$$ \begin{array}{lccccc} \lambda(\mathrm{nm}) & 405 & 435.8 & 480 & 520 & 577.7 \\ \hline \mathrm{KE}(\mathrm{J}) & 2.360 \times & 2.029 \times & 1.643 \times & 1.417 \times & 1.067 \times \\ & 10^{-19} & 10^{-19} & 10^{-19} & 10^{-19} & 10^{-19} \end{array} $$ (a) What is the lowest possible value of the principal quantum number \((n)\) when the angular momentum quantum number \((\ell)\) is \(1 ?\) (b) What are the possible values of the angular momentum quantum number ( \(\ell\) ) when the magnetic quantum number \(\left(m_{\ell}\right)\) is 0 , given than \(n \leq 4 ?\)

Short Answer

Expert verified
The lowest possible value for the principal quantum number when the angular momentum quantum number is \(1\) is \(2\). The possible values for the angular momentum quantum number when the magnetic quantum number is \(0\) and the principal quantum number is less than or equal to \(4\) are \(0, 1, 2, 3\).

Step by step solution

01

Understand Quantum Number Relationships

Understand quantum number relationships. The principal quantum number, n, can be any positive integer. The angular momentum quantum number, l, is dependent on n and can take on any integer value from 0 to n-1. The magnetic quantum number, ml, is related to l and can take integer values from -l to +l, including 0.
02

Find the quantum number n For part (a)

Notice that for l=1, according to the relationship between n and l, the smallest possible value for the principal quantum number n that allows l to be 1 would be n=2, as n has to be at least one unit larger than l.
03

Identify the Possible Values of l For part (b)

The magnetic quantum number ml is 0, and we know n is less than or equal to 4. Using the relationship between ml and l, there are four possible scenarios for l, which are l=0, l=1, l=2, and l=3. These are the possible values of angular momentum when ml is 0, and n is less than or equal to 4.

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