# 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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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 ...

#### Solution Summary

The solution provides and easy to follow explanation of the concepts and gives all mathematical steps wherever necessary.