
The Schrodinger wave equation for hydrogen atom is:
\[{\Psi _{2s}} = \dfrac{1}{{4\sqrt 2 }}{\left( {\dfrac{1}{{{a_0}}}} \right)^{\dfrac{3}{2}}}[2 - \dfrac{{{r_0}}}{{{a_0}}}]e{ - ^{\dfrac{r}{{{a_0}}}}}\]
Where ${a_0}$ is Bohr radius. If the radial node in 2s be at ${r_0}$ , then find r in terms of ${a_0}$
A.$\dfrac{{{a_0}}}{2}$
B.$2{a_0}$
C.$\sqrt {2{a_0}} $
D.$\dfrac{{{a_0}}}{{\sqrt 2 }}$
Answer
473.4k+ views
Hint: The radial node occurs where the radial component ${R_{nl}}(r)$ of the wave function goes to zero, therefore ${\Psi _{2s}}^2 = 0$ and At node, the radial node is at \[{r_0}\] , So \[[2 - \dfrac{{{r_0}}}{{{a_0}}}]\] = 0, then we can calculate r in terms of ${a_0}$
Complete step by step answer:
Given in the question,
The Schrodinger wave equation for hydrogen atom is
\[{\Psi _{2s}} = \dfrac{1}{{4\sqrt 2 }}{\left( {\dfrac{1}{{{a_0}}}} \right)^{\dfrac{3}{2}}}[2 - \dfrac{{{r_0}}}{{{a_0}}}]e{ - ^{\dfrac{r}{{{a_0}}}}}\]
The radial node occurs where the radial component ${R_{nl}}(r)$ of the wave function goes to zero. ${\Psi _{2s}}^2 = 0$ since there is no angular component \[{Y_I}^{ml}(\theta ,\emptyset )\] to a wave function for a spherical orbital \[(l = 0,ml = 0)\]
At node, the radial node is at \[{r_0}\]
0 = \[[2 - \dfrac{{{r_0}}}{{{a_0}}}]e{ - ^{\dfrac{r}{{{a_0}}}}}\]
Since \[e{ - ^{\dfrac{r}{{{a_0}}}}} \ne 0\] for r in between 0 and \[\infty \] (where nodes can occur), that can be divided out as well.
$\therefore 2 - \dfrac{{{r_0}}}{{{a_0}}} = 0$
$\dfrac{{{r_0}}}{{{a_0}}} = 2$
${r_0} = 2{a_0}$
Therefore, the correct answer is option (B).
Note: The wave function \[(\Psi )\] , is a mathematical function which is used to describe a quantum object. The wave function that describes an electron in an atom is a product between the radial wave function and the angular wave function. The radial wave function depends only on the distance from the nucleus and is represented by r.
A node occurs when a wave function changes signs, i.e. when its passes through zero. And a radial node occurs when a radial wave function passes through zero. An electron has the zero probability of being located at a node.
Complete step by step answer:
Given in the question,
The Schrodinger wave equation for hydrogen atom is
\[{\Psi _{2s}} = \dfrac{1}{{4\sqrt 2 }}{\left( {\dfrac{1}{{{a_0}}}} \right)^{\dfrac{3}{2}}}[2 - \dfrac{{{r_0}}}{{{a_0}}}]e{ - ^{\dfrac{r}{{{a_0}}}}}\]
The radial node occurs where the radial component ${R_{nl}}(r)$ of the wave function goes to zero. ${\Psi _{2s}}^2 = 0$ since there is no angular component \[{Y_I}^{ml}(\theta ,\emptyset )\] to a wave function for a spherical orbital \[(l = 0,ml = 0)\]
At node, the radial node is at \[{r_0}\]
0 = \[[2 - \dfrac{{{r_0}}}{{{a_0}}}]e{ - ^{\dfrac{r}{{{a_0}}}}}\]
Since \[e{ - ^{\dfrac{r}{{{a_0}}}}} \ne 0\] for r in between 0 and \[\infty \] (where nodes can occur), that can be divided out as well.
$\therefore 2 - \dfrac{{{r_0}}}{{{a_0}}} = 0$
$\dfrac{{{r_0}}}{{{a_0}}} = 2$
${r_0} = 2{a_0}$
Therefore, the correct answer is option (B).
Note: The wave function \[(\Psi )\] , is a mathematical function which is used to describe a quantum object. The wave function that describes an electron in an atom is a product between the radial wave function and the angular wave function. The radial wave function depends only on the distance from the nucleus and is represented by r.
A node occurs when a wave function changes signs, i.e. when its passes through zero. And a radial node occurs when a radial wave function passes through zero. An electron has the zero probability of being located at a node.
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