The orbital angular momentum of an electron in 3 s orbital is (x h)/(2 π). The value of x is ____________ (nearest integer)

Solution

Chemistry ma’am · 4 steps

Orbital angular momentum depends on l, not n, and every s orbital has l = 0.

1. Use the right formula. The orbital angular momentum of an electron is fixed by the azimuthal quantum number l alone. The principal quantum number n does not appear in it.

   L = √l(l+1) h/2π

2. Find l. For any s orbital, l = 0. This holds for 1s, 2s, 3s and every other s orbital, so the 3 in 3s does not matter here.
3. Put in l. L = √(0 × (0+1)) h/2π = 0. Comparing with xh/2π gives x = 0.
4. So. An electron in a 3s orbital has zero orbital angular momentum, so x = 0.

The common slip: 3. Using Bohr's formula nh/2π with n = 3 gives 3. That formula is for Bohr orbits; for an orbital you need l, which is 0 for s.

The idea to keep. For orbital angular momentum, use √l(l+1) h/2π and read l from the orbital letter. Every s orbital gives zero. Bohr's nh/2π belongs to the Bohr model, not to orbitals.

Answer from the NTA final answer key. Solution by PrepWiser.