For electrons in ' 2 s ' and ' 2 p ' orbitals, the orbital angular momentum values, respectively are:
Solution
<p>The orbital angular momentum for electrons in orbitals is given by the formula:</p>
<p>$ \text{Orbital angular momentum} = \sqrt{\ell(\ell+1)} \frac{\mathrm{h}}{2 \pi} $</p>
<p>For the 2s orbital, the quantum number $\ell$ is 0:</p>
<p>Orbital angular momentum = $\sqrt{0(0+1)} \frac{\mathrm{h}}{2 \pi} = 0$</p>
<p>For the 2p orbital, the quantum number $\ell$ is 1:</p>
<p>Orbital angular momentum = $\sqrt{1(1+1)} \frac{\mathrm{h}}{2 \pi} = \sqrt{2} \frac{\mathrm{h}}{2 \pi}$</p>
About this question
Subject: Chemistry · Chapter: Atomic Structure · Topic: Bohr's Model
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