Medium
For an electron with $n = 3$,there is only one radial node. The orbital angular momentum of the electron will be
- A$0$
- B$\sqrt{6} \frac{h}{2\pi}$
- C$\sqrt{2} \frac{h}{2\pi}$
- D$3 \left(\frac{h}{2\pi}\right)$
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Similar Questions
The number of radial nodes in a $2s$ orbital is equal to that in a ........ orbital.
Difficult
View SolutionMatch List-$I$ with List-$II$ :
List-$I$ (Quantum numbers) List-$II$ (Orbital) $A$. $n = 2, l = 1$ $I$. $3d$ $B$. $n = 4, l = 0$ $II$. $2p$ $C$. $n = 5, l = 3$ $III$. $4s$ $D$. $n = 3, l = 2$ $IV$. $5f$
Choose the correct answer from the options given below :
| List-$I$ (Quantum numbers) | List-$II$ (Orbital) |
| $A$. $n = 2, l = 1$ | $I$. $3d$ |
| $B$. $n = 4, l = 0$ | $II$. $2p$ |
| $C$. $n = 5, l = 3$ | $III$. $4s$ |
| $D$. $n = 3, l = 2$ | $IV$. $5f$ |
Choose the correct answer from the options given below :
MediumNEET 2026
View SolutionThe number of electrons in the outermost shell of an element with atomic number $15$ is . . . . . . .
Medium
View SolutionMatch List-$I$ with List-$II$ :
List-$I$ (Quantum number) List-$II$ (Orbital) $a. n=2, \ell=1$ $i. 2s$ $b. n=3, \ell=2$ $ii. 3s$ $c. n=3, \ell=0$ $iii. 2p$ $d. n=2, \ell=0$ $iv. 3d$
Choose the correct answer from the options given below :
| List-$I$ (Quantum number) | List-$II$ (Orbital) |
| $a. n=2, \ell=1$ | $i. 2s$ |
| $b. n=3, \ell=2$ | $ii. 3s$ |
| $c. n=3, \ell=0$ | $iii. 2p$ |
| $d. n=2, \ell=0$ | $iv. 3d$ |
Choose the correct answer from the options given below :
MediumNEET 2022
View SolutionWhat is the lowest value of $n$ that allows $g$ orbitals to exist?
Easy
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