See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

(A) The total lines obtained due to the splitting of a spectral line in the presence of magnetic effect is (2l + 1) as the presence of orbitals which have specific orientation in the presence of external field take up certain new orientation. The number of orbitals are equal to (2l + 1) and for each orbital one splitting takes place. (B) As we move away from the nucleus the difference in the energy levels become lesser and lesser hence. For 1st line of Lyman E =         2 n 2 1 1 n 1 – n 1 E E1 = E1     4 3 – 1st line; E2 = E1     9 8 –2nd line E3 = E1     16 15 –34rd line ; E4 = E1     25 24 – 4th line; E5 = E1     36 35 – 5th line; E2 – E1 = E1     4 3 – 9 8 = E1     36 5 E3 – E2 = E1     9 8 – 16 15 = E1     144 128 – 135 = E1     144 7 E4 – E3 = E1     16 15 – 25 24 = E1        16 25 25 15 – 24 16 = E1     400 9 We find that (E2 – E1) > (E3 – E2) > (E4 – E3) Hence the distance also become lesser and lesser. (C) The distance from the nucleus for maximum probability of finding electrons is 0.53 Å. This is not on the circumference of the orbital but is in the vicinity of the circumference.