Advertisements
Advertisements
Question
Calculate the total number of angular nodes and radial nodes present in 3p orbital.
Advertisements
Solution
The number of radial nodes is given by n – l – 1, where n is the principal quantum number, l is the azimuthal quantum number.
The number of angular nodes is given by n – l, where n is the principal quantum number, l is the azimuthal quantum number.
Here n = 3 and l = 1
Thus, angular nodes = 3 – 1 = 2 and radial node = 3 – 1 – 1 = 1.
APPEARS IN
RELATED QUESTIONS
Using s, p, d notations, describe the orbital with the following quantum numbers n = 1, l = 0.
Using s, p, d notations, describe the orbital with the following quantum numbers n = 3; l =1.
State Heisenberg uncertainty principle.
Give the names of quantum numbers.
Write orbital notations for the electron in orbitals with the following quantum numbers.
n = 4, l = 2
Write orbital notations for the electron in orbitals with the following quantum numbers.
n = 3, l = 2
Write condensed orbital notation of electronic configuration of the following element:
Carbon (Z = 6)
Write condensed orbital notation of electronic configuration of the following element:
Chlorine (Z = 17)
If n = 3, what are the quantum number l and m?
Write a note on ‘Principal Quantum number.
Which mineral among the following contains vanadium in it?
The designation of a subshell with n = 6 and l = 2 is ____________.
Which one of the following orders is CORRECT in case of energy of the given subshells?
P: n = 4; l = 3
Q: n = 5; I = 1
R: n = 5; l = 0
S: n = 4; l = 2
How many electrons in 19K have n = 3, l = 1?
How many electrons can fit in the orbital for which n = 4 and l = 2?
Which of the following options does not represent ground state electronic configuration of an atom?
The number of radial nodes for 3p orbital is ______.
Orbital angular momentum depends on ______.
The pair of ions having same electronic configuration is ______.
The electronic configuration of valence shell of Cu is 3d104s1 and not 3d94s2. How is this configuration explained?
