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
Verified
464.4k+ 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.
Recently Updated Pages
Master Class 9 Science: Engaging Questions & Answers for Success
Master Class 9 English: Engaging Questions & Answers for Success
Class 9 Question and Answer - Your Ultimate Solutions Guide
Master Class 9 Maths: Engaging Questions & Answers for Success
Master Class 9 General Knowledge: Engaging Questions & Answers for Success
Class 10 Question and Answer - Your Ultimate Solutions Guide
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
State and prove Bernoullis theorem class 11 physics CBSE
Pigmented layer in the eye is called as a Cornea b class 11 biology CBSE
10 examples of friction in our daily life
The lightest gas is A nitrogen B helium C oxygen D class 11 chemistry CBSE
State the laws of reflection of light