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Question: For 2s orbital, the nodal surface at a distance of: A. \[{a_0}\] from the nucleus B. \[1.5{a_0}\...

For 2s orbital, the nodal surface at a distance of:
A. a0{a_0} from the nucleus
B. 1.5a01.5{a_0} from the nucleus
C. 2a02{a_0} from the nucleus
D. none of these

Explanation

Solution

Nodal surface is the region where the probability of finding an electron is zero.
Schrodinger wave function for 2s- subshell can be represented as,
φ(2s)=142πa032(2−ra0)e(−r2a0)\varphi (2s) = \dfrac{1}{{4\sqrt {2\pi } a_0^{\dfrac{3}{2}}}}\left( {2 - \dfrac{r}{{{a_0}}}} \right){e^{\left( {\dfrac{{ - r}}{{2{a_0}}}} \right)}}
Where,
φ\varphi -wave function
a0{a_0}-Bohr radius
r-distance from the nucleus

Complete step by step answer:
As we know that Node is the region where the probability of finding an electron is zero. Nodes have two types as radial and angular nodes. Angular node is the plane that contains the x-,y- and z-axis where its probability of finding an electron is zero. Angular nodes can be identified by their azimuthal quantum number. For s-orbital there are no angular nodes since its azimuthal quantum number is 0.
The nodal surface which is radial nodes can be calculated by the formula n−l−1n - l - 1
Where,
n-principal quantum number. For 2s orbital, n=2n = 2
l- azimuthal quantum number. For s-orbital, l=0l = 0
thus, the number of radial nodes for 2s-subshell=2−0−1⇒1 = 2 - 0 - 1 \Rightarrow 1
Schrodinger wave function for 2s- subshell can be represented as,
φ(2s)=142πa032(2−ra0)e(−r2a0)\varphi (2s) = \dfrac{1}{{4\sqrt {2\pi } a_0^{\dfrac{3}{2}}}}\left( {2 - \dfrac{r}{{{a_0}}}} \right){e^{\left( {\dfrac{{ - r}}{{2{a_0}}}} \right)}}
Where,
φ\varphi -wave function
a0{a_0}-Bohr radius(distance)
r-distance from the nucleus
At the nodal surface,φ(2s)=0\varphi (2s) = 0. So when (2−ra0)=0\left( {2 - \dfrac{r}{{{a_0}}}} \right) = 0, then the whole wave function for 2s- subshell becomes zero.
Thus, for 2s-subshell,
(2−ra0)=0\left( {2 - \dfrac{r}{{{a_0}}}} \right) = 0
⇒2=ra0\Rightarrow 2 = \dfrac{r}{{{a_0}}}
⇒r=2a0\Rightarrow r = 2{a_0}
Thus, for 2s orbital, the nodal surface is at a distance of 2a02{a_0} from the nucleus.

So, the correct answer is Option C.

Note: At the nodal surface, the wave function will become zero since electron density is zero on the nodal surface. The radial node is the spherical surface at which the probability of finding an electron is zero.