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Q.When a photon corresponding to the third line of Paschen series of hydrogen spectrum is emitted, determine the change in angular momentum of the electron associated with the process.

Odisha ChseOdisha CHSE +2 Science Board Exam 2018Subjective· 2mImportance★★★★★
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Third Paschen line is 6 to 3; L = n(h/2pi), so delta L = (6-3)h/2pi = 3h/2pi = 3.16 x 10^-34 J s.

Step 1 — Identify the transition. The Paschen series consists of transitions ending at nf=3n_f = 3. Its lines in order are 4→34\to3 (first), 5→35\to3 (second), 6→36\to3 (third). So the emitting transition is ni=6→nf=3n_i = 6 \to n_f = 3.

Step 2 — Bohr's quantisation of angular momentum: Ln=nh2π=nℏL_n = n\dfrac{h}{2\pi} = n\hbar.

Step 3 — Change in angular momentum:

∣ΔL∣=∣L6−L3∣=(6−3)h2π=3h2π|\Delta L| = |L_6 - L_3| = (6-3)\dfrac{h}{2\pi} = 3\dfrac{h}{2\pi}.

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