The orbit and orbital angular momentum of an electron are $\dfrac{{3h}}{{2\pi }}{{ }}and{{ }}\sqrt {\dfrac{3}{2}} \dfrac{h}{\pi }$ respectively. The number of radial and angular nodes for the orbital in which the electron is present are respectively.
A.$0,2$
B.$2,0$
C.$1,2$
D.$2,2$
Answer
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Hint:Electrons in free space may carry quantized angular momentum along the direction of their movement or propagation called the orbital angular momentum which is the sum of angular momenta of each electron. Radial nodes are given by the formula $n - l - 1$ where l is the angular node.
Complete step by step answer:
We know that orbital angular momentum is the sum of angular momenta that the electrons carry along their direction of propagation in free space. The angular momentum of a body changes with change in its radius. It basically symbolises the revolution of electrons in a fixed orbit around the nucleus.
Orbit angular momentum is given by the formula $\dfrac{{nh}}{{2\pi }}$ where n is the orbit in which the electron is
And orbital angular momentum is calculated by $\sqrt {l(l + 1)} \dfrac{h}{{2\pi }}$ in which l is an integer.
Orbit angular momentum
$ = \dfrac{{nh}}{{2\pi }}$
Here, $\dfrac{{nh}}{{2\pi }} = \dfrac{{3h}}{{2\pi }}$
$ \Rightarrow n = 3$
Orbital angular momentum
$ = \sqrt {l(l + 1)} \dfrac{h}{{2\pi }}$
Here, $\sqrt {l(l + 1)} \dfrac{h}{{2\pi }} = \sqrt {\dfrac{3}{2}} \dfrac{h}{\pi }$
On squaring both sides,
$\dfrac{{l(l + 1)}}{4} = \dfrac{3}{2}$
$ \Rightarrow l = 2$
Radial nodes are spherical surfaces where the probability of finding an electron is zero. As the quantum number increases, the number of radial nodes also increases. The number of radial nodes is given by $n - l - 1$
Angular node is the plane passing through the nucleus. Angular node is also called the nodal plane. It is given by $l$.
For the given question,
Radial nodes $ = n - l - 1$
$ \Rightarrow 3 - 2 - 1$
$
= 0 \\
\\
$
Angular nodes = l
$
\Rightarrow l = 2 \\
\\
$
So, the correct option for the above question is (A) $0,2$
Note:
An orbital is the space or region where we find electrons placed around the nucleus. Orbital node is a point or plane where the electron density in an orbital is zero. These orbital nodes are bordered by orbital lobes which border a section of two or nodes.
Complete step by step answer:
We know that orbital angular momentum is the sum of angular momenta that the electrons carry along their direction of propagation in free space. The angular momentum of a body changes with change in its radius. It basically symbolises the revolution of electrons in a fixed orbit around the nucleus.
Orbit angular momentum is given by the formula $\dfrac{{nh}}{{2\pi }}$ where n is the orbit in which the electron is
And orbital angular momentum is calculated by $\sqrt {l(l + 1)} \dfrac{h}{{2\pi }}$ in which l is an integer.
Orbit angular momentum
$ = \dfrac{{nh}}{{2\pi }}$
Here, $\dfrac{{nh}}{{2\pi }} = \dfrac{{3h}}{{2\pi }}$
$ \Rightarrow n = 3$
Orbital angular momentum
$ = \sqrt {l(l + 1)} \dfrac{h}{{2\pi }}$
Here, $\sqrt {l(l + 1)} \dfrac{h}{{2\pi }} = \sqrt {\dfrac{3}{2}} \dfrac{h}{\pi }$
On squaring both sides,
$\dfrac{{l(l + 1)}}{4} = \dfrac{3}{2}$
$ \Rightarrow l = 2$
Radial nodes are spherical surfaces where the probability of finding an electron is zero. As the quantum number increases, the number of radial nodes also increases. The number of radial nodes is given by $n - l - 1$
Angular node is the plane passing through the nucleus. Angular node is also called the nodal plane. It is given by $l$.
For the given question,
Radial nodes $ = n - l - 1$
$ \Rightarrow 3 - 2 - 1$
$
= 0 \\
\\
$
Angular nodes = l
$
\Rightarrow l = 2 \\
\\
$
So, the correct option for the above question is (A) $0,2$
Note:
An orbital is the space or region where we find electrons placed around the nucleus. Orbital node is a point or plane where the electron density in an orbital is zero. These orbital nodes are bordered by orbital lobes which border a section of two or nodes.
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