Fe has the electron configuration [Ar]3d64s2. What is the magnitude of the orbital angular momentum for its most energetic elect
1 answer:
Answer : The magnitude of the orbital angular momentum for its most energetic electron is,
Explanation :
The formula used for orbital angular momentum is:
where,
L = orbital angular momentum
l = Azimuthal quantum number
As we are given the electronic configuration of Fe is,
Its most energetic electron will be for 3d electrons.
The value of azimuthal quantum number(l) of d orbital is, 2
That means, l = 2
Now put all the given values in the above formula, we get:
Therefore, the magnitude of the orbital angular momentum for its most energetic electron is,
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Answer:
2.40 x 10⁻¹³ C
Explanation:
= number of electrons = 6.25 x 10⁶
= charge on electron = - 1.6 x 10⁻¹⁹ C
= number of protons = 7.75 x 10⁶
= charge on proton = 1.6 x 10⁻¹⁹ C
Net charge is given as
Q =
+
Q = (- 1.6 x 10⁻¹⁹) (6.25 x 10⁶) + (1.6 x 10⁻¹⁹) (7.75 x 10⁶)
Q = 2.40 x 10⁻¹³ C
<h2>The rank is : 
.</h2>
We need to rank the objects according to the magnitude of their momentum.
We know momentum ,{ here m is mass and v is velocity }
Momentum of object A ,
Momentum of object B ,
Momentum of object C ,
Now, we can rank them in order of their magnitude of momentum .
.
Hence, this is the required solution.
Learn More :
Momentum