(Solved) - The magnitude of the orbital angular momentum in an excited state... (1 Answer) | Transtutors (2024)

FAQs

What is the equation for the magnitude of orbital angular momentum? ›

The angular momentum formula for any given orbital is expressed as L = l ( l + 1 ) ℏ , where is the magnitude of the angular momentum, is the angular momentum quantum number, and is the reduced Planck's constant (approximately 1.054571 x 10 − 34 Js).

Calculating angular momentum

If the object is a point particle of mass m, rotating with instantaneous speed v about an axis and at a fixed distance r from that axis (so that the particle is in a circular orbit), then I=mr2 and ω​ has magnitude ω=rv​. Hence L​ has magnitude L=mr2rv​=mrv in this special case.

The orbital angular momentum for an electron revolving in an orbit is given by √l(l+1)h2π.

The magnitude of the angular momentum for an electron with an angular momentum quantum number of 1 is given by square root of l times l plus 1 times Planck's constant divided by 2π.

p = m v . You can see from the equation that momentum is directly proportional to the object's mass (m) and velocity (v). Therefore, the greater an object's mass or the greater its velocity, the greater its momentum.

L=∑i(→li)z=∑iRiΔmivi=∑iRiΔmi(Riω)=ω∑iΔmi(Ri)2. L=Iω. This equation is analogous to the magnitude of the linear momentum p = mv. The direction of the angular momentum vector is directed along the axis of rotation given by the right-hand rule.

Formula to calculate angular momentum (L) = mvr, where m = mass, v = velocity, and r = radius.

$L = mvr$, where L is the angular momentum of the planet, m is the mass, v is the velocity and r is the radius from the centre that is the Sun. Hence, the correct answer is option (C). Kepler's Law states that the angular momentum of a planet remains constant as it moves in an orbit around the Sun.

The angular momentum of an electron by Bohr is given by mvr or nh/2π (where v is the velocity, n is the orbit in which the electron is revolving, m is mass of the electron, and r is the radius of the nth orbit).

What is orbit and orbital angular momentum? ›

We know that orbital angular momentum is the sum of angular momenta that the electrons carry along their direction of propagation in free space. The angular momentum of a body changes with change in its radius. It basically symbolises the revolution of electrons in a fixed orbit around the nucleus.

The magnitude of the orbital angular momentum of an electron is given by L=√5hπ.

Explanation: Orbital angular momentum depends on the value of l which is referred to as the azimuthal quantum number.

The magnitude of the total angular momentum can be found using the formula j ( j + 1 ) ℏ , where is the reduced Planck's constant. Therefore, the magnitude of the electron's total angular momentum is ( 5 / 2 ) ( 7 / 2 ) ℏ = 35 / 4 ℏ .

For p-orbital l=1; orbital angular momentum is given is =√l(l+1)h2π=√1(1+1)h2π=√2h2π=h√2π

In 2D, the cross product of two vectors is given by L = r * p * sin(theta), where theta is the angle between the vectors. In this case, the angle between the position vector r and the linear momentum vector p is 90 degrees, so sin(theta) is 1. Therefore, the magnitude of the angular momentum is L = m * d * g.

The angular momentum of an electron by Bohr is given by mvr or nh/2π (where v is the velocity, n is the orbit in which the electron is revolving, m is mass of the electron, and r is the radius of the nth orbit).