Answer:
(√ 20) hbar.
Explanation:
The first thing to do is to calculate the second principal quantum number, n from the formula for Calculating energy,E and this is done as given below;
Energy, E = 13.6 (1 / n^2 - 1/6).
(Where energy,E = 1240/ 7474).
Hence, when we solve the above equation for n, the principal quantum number = 5.
We know that the azimuthal quantum Number, l = (n- 1); 5 - 1 = 4.
Therefore, the maximum possible magnitude of the orbital angular momentum of the atom after emission = (√ 4 × 5) hbar.
the maximum possible magnitude of the orbital angular momentum of the atom after emission = (√ 20) hbar.
Note that hbar = 1.0545718 × 10-34 m2 kg / s.
