# The Schrodinger wave equation for hydrogen atom is

$\Psi \left( {{\text{radial}}} \right) = \dfrac{1}{{16\sqrt 4 }}{\left( {\dfrac{Z}{{{a_0}}}} \right)^{3/2}}\left[ {\left( {\sigma - 1} \right)\left( {{\sigma ^2} - 8\sigma + 12} \right)} \right]{e^{ - \sigma /2}}$

where ${a_0}$ and $Z$ are the constants in which answer can be expressed and $\sigma = \dfrac{{2Zr}}{{{a_0}}}$minimum and maximum position of radial nodes from nucleus are _______ respectively.

A.$\dfrac{{{a_0}}}{Z},\dfrac{{3{a_0}}}{Z}$

B.$\dfrac{{{a_0}}}{{2Z}},\dfrac{{{a_0}}}{Z}$

C.$\dfrac{{{a_0}}}{{2Z}},\dfrac{{3{a_0}}}{Z}$

D.$\dfrac{{{a_0}}}{{2Z}},\dfrac{{4{a_0}}}{Z}$

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**Hint:**To answer this question, you must recall the quantum mechanical model of an atom given by Irwin Schrodinger. According to Schrodinger, electrons are represented by an amplitude wave function notated by the Greek alphabet psi: $\Psi $.

**Complete step by step answer:**

We know that, at radial nodes, ${\Psi ^2} = 0$

Thus, we can say $\Psi = 0$

Now we are given the following relation in the question $\Psi \left( {{\text{radial}}} \right) = \dfrac{1}{{16\sqrt 4 }}{\left( {\dfrac{Z}{{{a_0}}}} \right)^{3/2}}\left[ {\left( {\sigma - 1} \right)\left( {{\sigma ^2} - 8\sigma + 12} \right)} \right]{e^{ - \sigma /2}}$

Solving for $\Psi = 0$,

We can write $\sigma - 1 = 0$and ${\sigma ^2} - 8\sigma + 12 = 0$

Solving, we get, $\sigma = 1$

Substituting this value into $\sigma = \dfrac{{2Zr}}{{{a_0}}}$, we can find the distance ${r_1}$as, $r = \dfrac{{{a_0}}}{{2Z}}$

Solving the equation, ${\sigma ^2} - 8\sigma + 12 = 0$, we get, $\left( {\sigma - 6} \right)\left( {\sigma - 2} \right) = 0$

Thus, we have $\sigma = 6$ and $\sigma = 2$

The values of $r$ for these values of $\sigma $are ${r_6} = \dfrac{{3{a_0}}}{Z}$and ${r_2} = \dfrac{{{a_0}}}{Z}$respectively.

**Thus, the correct answer is C.**

**Note:**

In the quantum mechanical model of an atom, the electron is believed to be in the form of a wave moving around the nucleus in 3-D space with constant energy. There are some regions around the nucleus where the probability of finding electrons is very high since they are well- defined quantized states that have a minimum possible energy and maximum stability in that region.

When we solve Schrodinger’s equation, the solution gives us the possible energy levels that can be occupied by the electrons and the corresponding wave functions of the electron in each of these energy levels. These quantized energy states and corresponding wave functions are characterized by a set of three quantum numbers ( $n,l,m$). The wave functions arise as a natural consequence in the solution of the Schrodinger equation.

When an electron is present in any energy state, the wave function corresponding to that energy state provides us all the information about that electron. A wave function is a mathematical function that depends on the coordinates of the electron in the atom. It does not carry any physical significance of its own. These wave functions are termed as atomic orbitals.