# Working with schrodinger atomic model - Quantum states

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According to the Schrodinger atomic model, each quantum state of an atom or ion can be labelled using two quantum numbers n and l.

Write down a formula for the energy of each of the quantum states of the hydrogen like boron ion B4+.

For a given value of n, state how the number of quantum states depends on the value of the second quantum number l.

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

Please see the attached document.

According to the Schrodinger atomic model, each quantum state of an atom or ion can be labelled using two quantum numbers n and l.

Write down a formula for the energy of each of the quantum states of the hydrogen like boron ion B4+.

For a given value of n, state how the number of quantum states depends on the value of the second quantum number l.

I have sent a detailed explanation on the last posting by you to the BM admin. I am adding it here also.

Let us consider the case of an ion with the charge of nucleus being Ze and an electron moving with a constant speed v along a circle of radius r with the center at the nucleus. The force acting on the electron is that due to Coulomb attraction and is equal to

F = Ze2/40r2

The acceleration of the electron is towards the center and has a magnitude v2/r. If m is the mass of the electron, from Newton's law we obtain

Ze2/40r2 = mv2/r

Using Bohr's angular momentum quantization rule for the value n, the Principal quantum number, we obtain both the velocity v, and the radius r as:

v = Ze2/20hn r = 0h2n2/ mZe2 ...(i)

We see that the allowed radii are proportional to n2. For each value of n, we have an allowed orbit. For n=1, we have the first orbit (smallest radius) , for n=2, we have the second orbit and so on.

The kinetic energy of the electron in the nth orbit is

K.E = mv2/2 = mZ2e4/8 02h2n2 ...(ii)

The potential energy of the atom is

P.E = -Ze2/40r = -mZ2 e4/402h2 n2...(iii)

We have taken the potential energy to be zero when the nucleus and the electron are widely separated. The total energy of the atom is

E = K.E+P.E = -mZ2e4/80 2h2n2 ...(iv)

Equations (i) to (iv) give various parameters of the atom when the electron is in the nth orbit .The atom is also said to be in the nth energy state in this case.

h = 6.625*10-34, 0 = 8.85418782 × 10-12 , e = 1.6*10-19 C and m = 9.1*10-31 Kg

En = -mZ2e4/80 2h2n2

= - Z2 9.1*10-31*(1.6*10-19)4 /[8*(8.85×10-12)2 *(6.625*10-34)2 ] n2

= -Z2 * 2.17*10^-18/ n2 Joules

1 Joule = 6.24150974 × 1018 electron volts

Thus En = -Z2 * 2.17*10^-18/ n2 Joules * 6.24150974 × 1018 electron volts

= - [Z2/n2] * 13.6 eV

The energy of any state depends on the value of Z (atomic number) and n.

For boron Z = 5.

Thus En = - [52/n2] * 13.6 eV = -13.6*25/n2 eV = -340/n2 eV

To put boron in the first excited state (ie, to remove one electron from it) we must supply 340 eV of energy. As the value of n increases, the electron is farther away from the nucleus and the energy required gets reduced.

• n, the principal quantum number. This is also known as the radial quantum number, and defines the distance of the electron from the nucleus in the Bohr model. n also describes the azimuthal angular momentum. n takes on integral values 1, 2, 3, ... .

• l, the azimuthal quantum number. In Sommerfeld's generalization of the Bohr model, the circular orbits of the electrons are replaced by elliptical orbits, and l describes the shape of the orbit. l takes on the integral values 0, 1, 2, ... , n-2, n-1. If n=1, l=0. l is sometimes called the reduced azimuthal quantum number, because the Sommerfeld formulation used a quantum number k, which equals l+1. k=0 corresponds to no angular momentum, or a radial orbit which takes the electron through the nucleus. This is unphysical, and is forbidden.

• m, the magnetic quantum number. m takes on the integral values -l , -(l-1), ..., -1, 0, 1, ..., (l-1), l. In Sommerfeld's formulation, m described the orientation of the ellipse. This is known as the magnetic quantum number because its effects are generally observed only under the influence of strong magnetic fields (which set a preferred spatial orientation).

• s, the spin quantum number. This describes the spin of the electron, and is either +1/2 or -1/2.

Thus for a given value of n, l will vary from 0 to n-1

If n = 5, l will have values 0, 1, 2, 3, 4 (five values)

The various states relating to different values of l are sometimes called sub-shells. A shell consists of all those states with the same value of n, the principal quantum number. A subshell groups all the states within one shell with the same value of , the orbital quantum number.

For a given value of the Principal quantum number, n, the possible values of l range from 0 to n-1; therefore, the n=1 shell only possesses an s subshell and can only take 2 electrons, the n=2 shell possesses an s and a p subshell and can take 8 electrons overall, the n=3 shell possesses s, p and d subshells and has a maximum of 18 electrons, and so on (generally speaking, the maximum number of electrons in the nth energy level is 2n2).

l value Letter Maximum number of electrons in shell

0 s 2

1 p 6

2 d 10

3 f 14

4 g 18

According to the Schrodinger atomic model, each quantum state of an atom or ion can be labelled using two quantum numbers n and l.

Write down a formula for the energy of each of the quantum states of the hydrogen like boron ion B4+.

For a given value of n, state how the number of quantum states depends on the value of the second quantum number l.

I have sent a detailed explanation on the last posting by you to the BM admin. I am adding it here also.

Let us consider the case of an ion with the charge of nucleus being Ze and an electron moving with a constant speed v along a circle of radius r with the center at the nucleus. The force acting on the electron is that due to Coulomb attraction and is equal to

F = Ze2/40r2

The acceleration of the electron is towards the center and has a magnitude v2/r. If m is the mass of the electron, from Newton's law we obtain

Ze2/40r2 = mv2/r

Using Bohr's angular momentum quantization rule for the value n, the Principal quantum number, we obtain both the velocity v, and the radius r as:

v = Ze2/20hn r = 0h2n2/ mZe2 ...(i)

We see that the allowed radii are proportional to n2. For each value of n, we have an allowed orbit. For n=1, we have the first orbit (smallest radius) , for n=2, we have the second orbit and so on.

The kinetic energy of the electron in the nth orbit is

K.E = mv2/2 = mZ2e4/8 02h2n2 ...(ii)

The potential energy of the atom is

P.E = -Ze2/40r = -mZ2 e4/402h2 n2...(iii)

We have taken the potential energy to be zero when the nucleus and the electron are widely separated. The total energy of the atom is

E = K.E+P.E = -mZ2e4/80 2h2n2 ...(iv)

Equations (i) to (iv) give various parameters of the atom when the electron is in the nth orbit .The atom is also said to be in the nth energy state in this case.

h = 6.625*10-34, 0 = 8.85418782 × 10-12 , e = 1.6*10-19 C and m = 9.1*10-31 Kg

En = -mZ2e4/80 2h2n2

= - Z2 9.1*10-31*(1.6*10-19)4 /[8*(8.85×10-12)2 *(6.625*10-34)2 ] n2

= -Z2 * 2.17*10^-18/ n2 Joules

1 Joule = 6.24150974 × 1018 electron volts

Thus En = -Z2 * 2.17*10^-18/ n2 Joules * 6.24150974 × 1018 electron volts

= - [Z2/n2] * 13.6 eV

The energy of any state depends on the value of Z (atomic number) and n.

For boron Z = 5.

Thus En = - [52/n2] * 13.6 eV = -13.6*25/n2 eV = -340/n2 eV

To put boron in the first excited state (ie, to remove one electron from it) we must supply 340 eV of energy. As the value of n increases, the electron is farther away from the nucleus and the energy required gets reduced.

? n, the principal quantum number. This is also known as the radial quantum number, and defines the distance of the electron from the nucleus in the Bohr model. n also describes the azimuthal angular momentum. n takes on integral values 1, 2, 3, ... .

? l, the azimuthal quantum number. In Sommerfeld's generalization of the Bohr model, the circular orbits of the electrons are replaced by elliptical orbits, and l describes the shape of the orbit. l takes on the integral values 0, 1, 2, ... , n-2, n-1. If n=1, l=0. l is sometimes called the reduced azimuthal quantum number, because the Sommerfeld formulation used a quantum number k, which equals l+1. k=0 corresponds to no angular momentum, or a radial orbit which takes the electron through the nucleus. This is unphysical, and is forbidden.

? m, the magnetic quantum number. m takes on the integral values -l , -(l-1), ..., -1, 0, 1, ..., (l-1), l. In Sommerfeld's formulation, m described the orientation of the ellipse. This is known as the magnetic quantum number because its effects are generally observed only under the influence of strong magnetic fields (which set a preferred spatial orientation).

? s, the spin quantum number. This describes the spin of the electron, and is either +1/2 or -1/2.

Thus for a given value of n, l will vary from 0 to n-1

If n = 5, l will have values 0, 1, 2, 3, 4 (five values)

The various states relating to different values of l are sometimes called sub-shells. A shell consists of all those states with the same value of n, the principal quantum number. A subshell groups all the states within one shell with the same value of , the orbital quantum number.

For a given value of the Principal quantum number, n, the possible values of l range from 0 to n-1; therefore, the n=1 shell only possesses an s subshell and can only take 2 electrons, the n=2 shell possesses an s and a p subshell and can take 8 electrons overall, the n=3 shell possesses s, p and d subshells and has a maximum of 18 electrons, and so on (generally speaking, the maximum number of electrons in the nth energy level is 2n2).

l value Letter Maximum number of electrons in shell

0 s 2

1 p 6

2 d 10

3 f 14

4 g 18

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