The angular momentum of an electron in an orbit having the highest magnetic quantum number 2, is:

\frac{1.5 h}{π}

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\frac{h}{π}

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\frac{h}{2 π}

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\frac{2 h}{π}

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Solution

The correct option is A $\frac{1.5 h}{π}$
Given, the orbit has its highest magnetic quantum number $m_{l}$ = 2
So, the principal quantum number $n$ = 3
We know, angular momentum $= \frac{n h}{2 π} = \frac{3 h}{2 π} = \frac{1.5 h}{π}$

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Question 21

Which of the following statements concerning the quantum numbers are correct ?

(a) The angular momentum quantum number determines the three-dimensional shape of the orbital

(b) The principal quantum number determines the orientation and energy of the orbital

(d). Spin quantum number of an electron determines the orientation of the spin of electron relative to the chosen axis.

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