Question

A) The quantum state of a particle can be specified by giving a complete set of quantum numbers (n,l, ml,ms). How many different quantum states are possible if the principal quantum number is n = 5?

To find the total number of allowed states, first write down the allowed orbital quantum numbers l, and then write down the number of allowed values of ml for each orbital quantum number. Sum these quantities, and then multiply by 2 to account for the two possible orientations of spin.

B) What is the maximum angular momentum Lmax that an electron with principal quantum number n= 3 can have?

Express your answer in units of ?. (You don't need to enter the ?, it is in the units field for you.)

Answer 1

For n=5, there are 50 different quantum states possible.

Step-by-step explanation:

To find the total number of allowed quantum states for the principal quantum number n=5, we need to find the allowed values of the orbital quantum number l and the number of allowed values for the magnetic quantum number ml for each orbital quantum number.

For n=5, the allowed values of l are 0, 1, 2, 3, and 4. The number of allowed values for ml depends on the value of l. For l=0, ml can only be 0. For l=1, ml can be -1, 0, or 1. For l=2, ml can be -2, -1, 0, 1, or 2. For l=3, ml can be -3, -2, -1, 0, 1, 2, or 3. For l=4, ml can be -4, -3, -2, -1, 0, 1, 2, 3, or 4.

Summing up the number of allowed values for ml for each l, we get 1 + 3 + 5 + 7 + 9 = 25. Multiplying this by 2 to account for the two possible orientations of spin, we find that there are 50 different quantum states possible for n=5.

Answer 2

The total number of different quantum states for a particle with a principal quantum number of n=5 is 50. The maximum angular momentum Lmax for an electron with principal quantum number n=3 is √6h/(2π).

Step-by-step explanation:

The number of different quantum states can be found by determining the values of l and ml for each value of n. For a principal quantum number of n=5, the possible values of l are 0, 1, 2, 3, and 4. For each value of l, the number of possible values of ml is given by 2l+1. So, when l=0, there is 1 value of ml. When l=1, there are 3 values of ml. When l=2, there are 5 values of ml. When l=3, there are 7 values of ml. When l=4, there are 9 values of ml. Therefore, the total number of possible quantum states is 1+3+5+7+9 = 25. Multiplying by 2 to account for the two possible spin orientations, the total number of different quantum states is 2 x 25 = 50.

The maximum angular momentum Lmax for an electron with a principal quantum number of n=3 can be determined by using the formula L = √l(l+1)h/(2π), where h is Planck's constant. For n=3, the possible values of l are 0, 1, and 2. Plugging in these values into the formula, we get L(0)=0, L(1)= √2h/(2π), and L(2)=√6h/(2π) respectively. Therefore, the maximum angular momentum Lmax for an electron with n=3 is Lmax=√6h/(2π).