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,
You might be interested in
The rod has a mass of m = 4.4 kg and a length of L = 0.4 m.
Its polar moment of inertia is
J = (mL²)/12
= (1/12) * [(4.4 kg)*(0.4 m)²]
= 0.05867 kg-m²
The mass of the bullet is 0.3 g.
If its velocity is v m/s, then its linear momentum is
P = (0.3 x 10⁻³ kg)*(v m/s)
Its linear momentum perpendicular to the rod is
P*sin(60°) = 2.5981 x 10⁻⁴ v (kg-m)/s
The angular momentum about the center of the rod when the bullet strikes is
T = (2.5981 x 10⁻⁴ v (kg-m)/s)*(0.2 m) = 5.1962 x 10⁻⁵ v (kg-m²)/s
Because the bullet lodges into the end of the rod, the combined polar moment of inertia is
J + (0.3 x 10⁻³ kg)*(0.2 m)² = 0.05867 + 1.2 x 10⁻⁵ = 0.0587 kg-m²
The initial angular velocity is ω = 17 rad/s.
Because angular momentum is conserved, therefore
5.1962 x 10⁻⁵ v (kg-m²)/s = (0.0587 kg-m²)*(17 rad/s)
v = 19204 m/s
Answer: 19204 m/s
Its polar moment of inertia is
J = (mL²)/12
= (1/12) * [(4.4 kg)*(0.4 m)²]
= 0.05867 kg-m²
The mass of the bullet is 0.3 g.
If its velocity is v m/s, then its linear momentum is
P = (0.3 x 10⁻³ kg)*(v m/s)
Its linear momentum perpendicular to the rod is
P*sin(60°) = 2.5981 x 10⁻⁴ v (kg-m)/s
The angular momentum about the center of the rod when the bullet strikes is
T = (2.5981 x 10⁻⁴ v (kg-m)/s)*(0.2 m) = 5.1962 x 10⁻⁵ v (kg-m²)/s
Because the bullet lodges into the end of the rod, the combined polar moment of inertia is
J + (0.3 x 10⁻³ kg)*(0.2 m)² = 0.05867 + 1.2 x 10⁻⁵ = 0.0587 kg-m²
The initial angular velocity is ω = 17 rad/s.
Because angular momentum is conserved, therefore
5.1962 x 10⁻⁵ v (kg-m²)/s = (0.0587 kg-m²)*(17 rad/s)
v = 19204 m/s
Answer: 19204 m/s