For electron in '2s' and '2p' orbitals, the orbital angular momentum values, respectively are :
Show Hint
- \( \sqrt{2} \frac{h}{2\pi} \) and 0
- \( \frac{h}{2\pi} \) and \( \sqrt{2} \frac{h}{2\pi} \)
- 0 and \( \sqrt{6} \frac{h}{2\pi} \)
- 0 and \( \sqrt{2} \frac{h}{2\pi} \)
The Correct Option is D
Approach Solution - 1
The orbital angular momentum of an electron in an atom is given by the formula: \[ L = \sqrt{l(l+1)} \frac{h}{2\pi} = \sqrt{l(l+1)} \hbar \] where \( l \) is the azimuthal quantum number (also known as the orbital angular momentum quantum number), and \( h \) is Planck's constant, with \( \hbar = \frac{h}{2\pi} \) being the reduced Planck constant.
For a '2s' orbital, the principal quantum number \( n = 2 \), and for an 's' orbital, the azimuthal quantum number \( l = 0 \). Substituting \( l = 0 \) into the formula for orbital angular momentum: \[ L_{2s} = \sqrt{0(0+1)} \frac{h}{2\pi} = \sqrt{0} \frac{h}{2\pi} = 0 \] So, the orbital angular momentum for an electron in a 2s orbital is 0. For a '2p' orbital, the principal quantum number \( n = 2 \), and for a 'p' orbital, the azimuthal quantum number \( l = 1 \). Substituting \( l = 1 \) into the formula for orbital angular momentum: \[ L_{2p} = \sqrt{1(1+1)} \frac{h}{2\pi} = \sqrt{1(2)} \frac{h}{2\pi} = \sqrt{2} \frac{h}{2\pi} \] So, the orbital angular momentum for an electron in a 2p orbital is \( \sqrt{2} \frac{h}{2\pi} \). The orbital angular momentum values for electrons in '2s' and '2p' orbitals are 0 and \( \sqrt{2} \frac{h}{2\pi} \) respectively.
This corresponds to option (4).
Approach Solution -2
The orbital angular momentum of an electron in an atomic orbital is given by the formula:
\(L = \sqrt{l(l+1)} \frac{h}{2\pi}\)
where:
- \(L\) is the angular momentum.
- \(l\) is the azimuthal quantum number (orbital quantum number).
- \(h\) is Planck's constant.
The azimuthal quantum number \(l\) determines the subshell of the electron:
- For an \(s\) orbital, \(l = 0\)
- For a \(p\) orbital, \(l = 1\)
Now, calculate the orbital angular momentum for:
- 2s orbital:
Since \(l = 0\)
\(L = \sqrt{0(0+1)} \frac{h}{2\pi} = 0\)
- 2p orbital:
Since \(l = 1\)
\(L = \sqrt{1(1+1)} \frac{h}{2\pi} = \sqrt{2} \frac{h}{2\pi}\)
Therefore, for electrons in 2s and 2p orbitals, the orbital angular momentum values are 0 and \(\sqrt{2} \frac{h}{2\pi}\) respectively.
So, the correct answer is: 0 and \(\sqrt{2} \frac{h}{2\pi}\).
Top JEE Main Chemistry Questions
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- At equilibrium the concentrations of and in a sealed vessel at . What will be for the reaction
- Find out the major products from the following reactions
Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
- The correct order of atomic radii is:
Top JEE Main Atomic Structure Questions
- Choose the correct option having all the elements with \( d^{10} \) electronic configuration from the following:
- The correct set of four quantum numbers for the valence electron of rubidium atom (Z = 37) is:
- $\textbf{Match List I with List II}$
$\begin{array}{|c|c|c|} \hline \text{List I (Spectral Series for Hydrogen)} & \text{List II (Spectral Region/Higher Energy State)} \\ \hline \text{A. Lyman} & \text{I. Infrared region} \\ \text{B. Balmer} & \text{II. UV region} \\ \text{C. Paschen} & \text{III. Infrared region} \\ \text{D. Pfund} & \text{IV. Visible region} \\ \hline \end{array}$
$\text{Choose the correct answer from the options given below :-}$ - Number of spectral lines obtained in He+ spectra, when an electron makes transition from fifth excited state to first excited state will be __________.
- Which of the following electronic configurations would be associated with the highest magnetic moment?
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.