Question

Angular momentum for the p-shell electron is:
(A) $\dfrac{{{\text{3h}}}}{{{\pi }}}$
(B) Zero
(C) $\dfrac{{\sqrt 2 {\text{h}}}}{{{{2\pi }}}}$
(D) None of these

Answer 1

The angular momentum of the electron arises due to the revolution of the electrons around the nucleus in their own shells and as per the Bohr’s Theory, the angular momentum is an integral multiple of the planck's constant divided by ${{2\pi }}$ .
Formula Used: Angular momentum
${\text{mvr = }}\sqrt {l\left( {l + 1} \right)} \dfrac{{\text{h}}}{{{{2\pi }}}}$ .

Complete step by step solution
The angular momentum of the electrons is represented by ${\text{L}}$ and is equal to $\sqrt {l\left( {l + 1} \right)} \dfrac{{\text{h}}}{{{{2\pi }}}}$ , where “l” is the azimuthal quantum number. Now, the azimuthal quantum number has different value for the different sub-shells, such as, for the “s” subshell the value is 0, for the “p” subshell, the value is 1, for the “f” subshell, the value is 2, and so on.
Putting the value of the azimuthal quantum number in the equation we get,
${\text{mvr = }}\sqrt {l\left( {l + 1} \right)} \dfrac{{{h}}}{{{{2\pi }}}} = \sqrt 2 \dfrac{{{h}}}{{{{2\pi }}}}$
Hence, the correct answer is option C.

Notes
The Bohr’s atomic model laid down different rules for the arrangement of the electrons in their shells around the nucleus and according to his theory, the values of the angular momentum of the electrons revolving around the nucleus is quantized and the electrons move only in those orbits where the angular momentum of the electrons is an integral multiple of $\dfrac{{\text{h}}}{{{{2\pi }}}}$ .
The Heisenberg’s Equation relates the change in the momentum of the electrons and the position of the electrons and states that it is impossible to determine the position and the momentum of the electron simultaneously and that their product is given by,
${{\Delta x \times \Delta p}} \geqslant \dfrac{{\text{h}}}{{{{4\pi }}}}$ , where ${{\Delta x}}$ is the change in Position and ${{\Delta p}}$ is the change in momentum.