Suppose that in an alternate universe, the possible values of l are the integer values from 0 to n (instead of 0 to nâ1). assuming no other differences between this universe and ours, how many orbitals would exist in each level in the alternate universe?

    level     possible values of ℓ     possible values of m     number of orbitals

    n = 1        0, 1                                 -1, 0, 1                                      3

    n = 2       0, 1, 2                             -2, - 1, 0, +1, +2                       5

    n = 3       0, 1, 2, 3                         -3, -2, -1, 0, +1, +2, +3            7    

    n = 4        ...                                    ...                                            9

    n = 5       ...                                    ...                                            11

Explanation:

1. This question is about the quantum numbers: n, ℓ, m, s.

2. n is the principal quantum number, i.e. amin energy level. It can be any positive integer: 1, 2, 3, 4, 5, 6, 7 ...

3. ℓ is the azimuthal quantum number. The possible values of ℓ are from 0 to n - 1.

4. It is stated that in the alternate universe the possible values of ℓ are from 0 to n, instead of o to  n - 1, and that all the other rules are still valid.

5. Since, two electrons in the same atom cannot have the same set of four quantum numbers, and the spin number can only be +1/2 or -1/2, the number of possible orbitals is equal to the amount  of possible ℓ numbers.

That implies that you can build the table with the possible number of orbitgals in each level:

    level     possible values of ℓ     possible values of m     number of orbitals

    n = 1        0, 1                                 -1, 0, 1                                      3

    n = 2       0, 1, 2                             -2, - 1, 0, +1, +2                       5

    n = 3       0, 1, 2, 3                         -3, -2, -1, 0, +1, +2, +3             7    

    n = 4        ...                                    ...                                            9

    n = 5       ...                                    ...                                            11

   n = 6        ...                                    ...                                            13

   n = 7        ...                                    ...                                            15

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s

no sign of dia of fuel cell