# Americium Bohr model

In the americium Bohr model, the nucleus comprises 95 protons and 148 neutrons. Surrounding this nucleus are seven electron shells, accommodating a total of 95 electrons.

To draw the americium Bohr model, represent the 95 protons, 148 neutrons, and 95 electrons. Begin by sketching the nucleus, and then draw the seven electron shells. The first six shells should contain 2, 8, 18, 32, 25, and 8 electrons, respectively, while the seventh shell holds the remaining 2 electrons.

Contents

## Steps

### Write protons, neutrons, and electrons of americium atom

Americium has 95 protons, 148 neutrons, and 95 electrons.

### Draw nucleus of americium atom

The nucleus of an americium atom contains 95 protons and 148 neutrons. So draw the nucleus of americium atom as follows:

Now in the next step, draw the 1st electron shell and start marking electrons.

### Draw 1st electron shell

Remember that we have a total of 95 electrons.

The 1st electron shell (containing s subshell) can hold up to a maximum of 2 electrons. So draw the 1st electron shell as follows:

In the above image, 1 represents the 1st electron shell that contains 1s subshell. And the green color represents the number of electrons in that subshell. This means that the 1st electron shell has a total of 2 electrons.

Since we have already used 2 electrons in the 1st electron shell, now we have 95 – 2 = 93 electrons left. So in the next step, we have to draw the 2nd electron shell.

### Draw 2nd electron shell

The 2nd electron shell (containing s subshell and p subshell) can hold up to a maximum of 8 electrons. So draw the 2nd electron shell as follows:

In the above image, 2 represents the 2nd electron shell that contains 2s and 2p subshells. And the green and orange color represents the number of electrons in that subshell. This means that the 2nd electron shell has a total of 8 electrons.

Now we have already used 10 electrons in 1st and 2nd electron shells, so we have 95 – 10 = 85 electrons left. So in the next step, we have to draw the 3rd electron shell.

### Draw 3rd electron shell

The 3rd electron shell (containing s subshell, p subshell, and d subshell) can hold up to a maximum of 18 electrons. So draw the 3rd electron shell as follows:

In the above image, 3 represents the 3rd electron shell that contains 3s, 3p, and 3d subshells. And the green, orange, and pink color represents the number of electrons in that subshell. This means that the 3rd electron shell has a total of 18 electrons.

Now we have already used 28 electrons in 1st, 2nd, and 3rd electron shells, so we have 95 – 28 = 67 electrons left. So in the next step, we have to draw the 4th electron shell.

### Draw 4th electron shell

The 4th electron shell (containing s subshell, p subshell, d subshell, and f subshell) can hold up to a maximum of 32 electrons. So draw the 4th electron shell as follows:

In the above image, 4 represents the 4th electron shell that contains 4s, 4p, 4d, and 4f subshells. And the green, orange, pink, and blue color represents the number of electrons in that subshell. This means that the 4th electron shell has a total of 32 electrons.

Now we have already used 60 electrons in 1st, 2nd, 3rd, and 4th electron shells, so we have 95 – 60 = 35 electrons left. So in the next step, we have to draw the 5th electron shell.

### Draw 5th electron shell

The 5th electron shell can hold up to a maximum of 50 electrons. So draw the 5th electron shell as follows:

In the above image, 5 represents the 5th electron shell that contains 5s, 5p, 5d, and 5f subshells. And the green, orange, pink, and blue color represents the number of electrons in that subshell. This means that the 5th electron shell has a total of 25 electrons.

In the 5th electron shell, the 5f subshell has only seven electrons. This is because according to the aufbau principle, the 6s subshell is filled first and then 4f, 5d, 6pâ€¦ and so on.

Now we have already used 85 electrons in 1st, 2nd, 3rd, 4th, and 5th electron shells, so we have 95 – 85 = 10 electrons left. So in the next step, we have to draw the 5th electron shell.

Still have a question: Why does the 5f subshell have only seven electrons?

Letâ€™s draw the orbital diagram of americium, and find the answer to the above question.

Answer: Once the 7s subshell is completely filled, then only we can mark electrons in the 5f subshell. Thatâ€™s why in the 5th electron shell, the 5f subshell has only seven electrons.

### Draw 6th electron shell

The 6th electron shell can hold up to a maximum of 72 electrons. So draw the 6th electron shell as follows:

In the above image, 6 represents the 6th electron shell that contains 6s and 6p subshells. And the green and orange color represents the number of electrons in that subshell. This means that the 6th electron shell has a total of 8 electrons.

The 6th electron shell contains only 6s and 6p subshells, and not 6d and 6f subshells. This is because according to the aufbau principle, the 7s subshell is filled first and then 5f, 6d, 7pâ€¦ and so on.

Now we have already used 93 electrons in 1st, 2nd, 3rd, 4th, 5th, and 6th electron shells, so we have 95 – 93 = 2 electrons left. So in the next step, we have to draw the 7th electron shell.

### Draw 7th electron shell

The 7th electron shell can hold up to a maximum of 98 electrons. So draw the 7th electron shell as follows:

In the above image, 7 represents the 7th electron shell that contains 7s subshell. And the green color represents the number of electrons in that subshell. This means that the 7th electron shell has a total of 2 electrons.

Thatâ€™s it! This is the final Bohr model of americium atom as we have used all 95 electrons: 2 electrons in the 1st electron shell, 8 electrons in the 2nd electron shell, 18 electrons in the 3rd electron shell, 32 electrons in the 4th electron shell, 25 electrons in the 5th electron shell, 8 electrons in the 6th electron shell, and 2 electrons in the 7th electron shell.