
Angular momentum for p−shell electron is:
A.\[\dfrac{3h}{\pi }\]
B.zero
C.\[\dfrac{\sqrt{2h}}{2\pi }\]
D.none of these
Answer
497.4k+ views
Hint: The atom can be described in terms of quantum numbers that give the total orbital angular momentum and total spin angular momentum of a given state for atoms in the first three rows and the first two columns of the periodic table.
Complete answer:
Electrons are organized into sets called shells based on their energies (labeled by the principal quantum number, n). In general, the higher a shell's energy, the farther it is (on average) from the nucleus. Shells do not have set distances from the nucleus, but an electron in a higher-energy shell will spend more time away from the nucleus than an electron in a lower-energy shell.
Shells are further subdivided into electron subsets known as subshells. The first shell has only one subshell, the second shell has two, the third shell has three, and so on. Each shell's subshells are labeled with the letters s, p, d, and f in that order.
Each p orbital is made up of two sections known as lobes that are located on either side of the plane that runs through the nucleus. The three p orbitals differ in the orientation of the lobes but are identical in size, shape, and energy.
For a \[p\]-shell electron, the orbital angular momentum is \[\dfrac{\left( \sqrt{l+\left( l+1) \right)} \right)h}{2\pi }\] where \[l\] is an integer The possible values \[l\] are determined by the individual \[l\] values and orbital orientations of all the electrons in the atom.
For \[p\]-shell,\[l=1\],
Thus, \[\dfrac{\left( \sqrt{1+\left( 1+1) \right)} \right)h}{2\pi }\]=\[\dfrac{\sqrt{2h}}{2\pi }\]
So, the answer is option C: \[\dfrac{\sqrt{2h}}{2\pi }\]
Note:
The letters s, p, d, and f stand for sharp, primary, diffuse, and fundamental, respectively. The letters and words refer to the visual impression left by the fine structure of the spectral lines as a result of the first relativistic corrections, specifically the spin-orbital interaction.
Complete answer:
Electrons are organized into sets called shells based on their energies (labeled by the principal quantum number, n). In general, the higher a shell's energy, the farther it is (on average) from the nucleus. Shells do not have set distances from the nucleus, but an electron in a higher-energy shell will spend more time away from the nucleus than an electron in a lower-energy shell.
Shells are further subdivided into electron subsets known as subshells. The first shell has only one subshell, the second shell has two, the third shell has three, and so on. Each shell's subshells are labeled with the letters s, p, d, and f in that order.
Each p orbital is made up of two sections known as lobes that are located on either side of the plane that runs through the nucleus. The three p orbitals differ in the orientation of the lobes but are identical in size, shape, and energy.
For a \[p\]-shell electron, the orbital angular momentum is \[\dfrac{\left( \sqrt{l+\left( l+1) \right)} \right)h}{2\pi }\] where \[l\] is an integer The possible values \[l\] are determined by the individual \[l\] values and orbital orientations of all the electrons in the atom.
For \[p\]-shell,\[l=1\],
Thus, \[\dfrac{\left( \sqrt{1+\left( 1+1) \right)} \right)h}{2\pi }\]=\[\dfrac{\sqrt{2h}}{2\pi }\]
So, the answer is option C: \[\dfrac{\sqrt{2h}}{2\pi }\]
Note:
The letters s, p, d, and f stand for sharp, primary, diffuse, and fundamental, respectively. The letters and words refer to the visual impression left by the fine structure of the spectral lines as a result of the first relativistic corrections, specifically the spin-orbital interaction.
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