Among the following electrons, which has highest energy? a. n = 3, l = 2, m = 0, s = + 1/2 b. n = 4, l = 0. m= 0 , s = - 1/2
Ans is A... but how A? Why nOt B?
To determine which electron has the highest energy, we need to consider the value of the principal quantum number (n). The higher the value of n, the higher the energy of the electron.
In option A, the electron has the quantum numbers n = 3, l = 2, m = 0, s = +1/2.
In option B, the electron has the quantum numbers n = 4, l = 0, m = 0, s = -1/2.
Comparing the values of n, we see that option B has a higher value of n (n = 4), while option A has a lower value of n (n = 3). Therefore, at first glance, it might seem that option B has the higher energy electron.
However, it's important to remember that the energy of an electron also depends on the value of l (the azimuthal quantum number) and not just the principal quantum number. In this case, l is 2 for option A and 0 for option B.
According to the filling order of electron energy levels, electrons with higher values of l (orbitals with higher subshells) have higher energy within the same value of n.
Since option A has a higher value of l (l = 2) compared to option B (l = 0), even though its value of n is lower, the electron in option A actually has higher energy.
Explanation: The energy of electrons within a shell depends on both the principal quantum number (n) and the azimuthal quantum number (l). Electrons with higher values of l have higher energy within the same value of n. Therefore, even though option B has a higher value of n, option A has a higher value of l, making its electron higher in energy.
Look at the order of filling. That is
1s, 2s, 2p, 3s, 3p, 4s, 3d etc.
So if n = 3 and l = 2, the electrons must be a 3d electron. For n = 4 and l = 0, that's a 4s electron. The 4s is lower energy than the 3d. I think B.