Question

Determine the angular momentum of an electron if $l\, = \,7$.

Answer 1

Hint
While studying about Quantum numbers, you must have come across the angular momentum quantum number $\left( l \right)$ in the chapter. There must have been the formula using which we can find the angular momentum using the angular momentum quantum number. Use that formula to find the correct answer.

Complete step by step answer
As we told in the hint, we need to use the formula which gives us the value of angular momentum using the value of angular momentum quantum number.
The formula that we will use is:
Angular momentum $\left( L \right)\, = \,\sqrt {l\left( {l + 1} \right)} \dfrac{h}{{2\pi }}$
Where, $h$ is the Planck’s Constant whose value is known to us as $6.626\, \times \,{10^{ - 34}}\,J\,H{z^{ - 1}}$
$l$ is the angular momentum quantum number, whose value is given to us in the question as $7$
So, we will be using the formula we told above and the value of $l$ that is given in the question to solve the question.
$L\, = \,\sqrt {l\left( {l + 1} \right)} \dfrac{h}{{2\pi }}$
Now we’ll substitute the value of $l\,\left( { = \,7} \right)$ as given in the question.
$L\, = \,\sqrt {7\left( {7 + 1} \right)} \,\dfrac{h}{{2\pi }}$
$L\, = \,\sqrt {56} \dfrac{h}{{2\pi }}$
Now, we will take $\sqrt 4 $ out from the RHS:
$L\, = \,2\sqrt {14} \dfrac{h}{{2\pi }}$
So, we can see that this is the answer to the question.

Note
This is an easy and marks-gaining topic. You should remember such formulae of various quantum numbers by heart. Many students get confused in the formula about whether it contains $l$ or $n$ . Always remember that the formula for angular momentum has $l$ in it. You can further put the value of Planck’s constant if the required answer is in numerical values, but mostly you’ll see the answer as this only. Try to avoid making calculation mistakes if numerical value is asked, carelessness can lose marks easily.