Question

3. Bohr atom. A hydrogen atom goes from the n = 4 state to the n=1 state, either all at once or via a series of intermediate downward jumps. a. How many different ways can this (these) transition(s) take place? Show each case in an energy level diagram. b. For each case, describe how many photon(s) are emitted, and of what wavelength(s)

Answer 1

3(a)

to u . to n=1 - one trannition ch (3) n to n=2 and then n:2 2 to na lii & 5 I 3s nzu to n=3 and then n=3 to nalan =4 to n=3 and then n=3 to n=2 and then n=2 to n=1 civo total u transitions can be possible n=4 n -u to n=2 ns1 AN ti [nu to n=2 & n=2 to nai] for i rase? nisu, ng=1 Only 1 Photon E= -13° (minst ]=-(- -) 5+36 (76-'] = -13,6 x 14 15 E = 12-15ev = 12.75x16x 1019 - f = - f = 2 - - 6.63 x 15 34 J.S. If 3.576x10's uz]

(b)​​​​
da $ - Booking Hslotm casei fromn=4to and then Ez-130 [ ] =-13.0(loca E, 2.550 n=2 to 3 - 2 photons , 6==13067 ) 5-3-4[42] F zlozer 2 255 41-6x1017 ; f=€ - 10:281-6x10') - 6.636x1034 6.6% 1034 If 6:15 x 104 117 , f = 2.45x105 412) 3x108 ult ç = 3x10 6.15 x 2014 , 15 g 3.45 's To 488m] / [A= 1.22x10 "m Case in fromn=4 to 3 and then > 13.45-+] 4.136 ( 7 E = o.ober n=3 to na 2 photory E.13"[ +) [E = 12:08 ev]

f 0 -6631: O KTÓW IfE - 12.08 x 16 x 1619 - 6.636 x 1534 f = 1.59x109H7 , f = 2,9x10's ] C 3 = 3x108 1.59x1014 * 10% 2.9x10's f - (d=1.88x100m ] [1=1,03x10m gare civ 3 photons gets emitted nou bo 3 and then n=3 to 2 and then na 2 ton - only need to calculate n=3 to 2. since we already have no u to 3 and n=2 tol from above cases i n=3 toz E = -130[$7 9-1346) LE 21.88ev 16.636x1094 $ € F.4.55x10947 C = 3vto 4.55x104 6.59 16m

Where nf is final state and ni is initial state. E is the energy, f is frequency and $\lambda$ is wavelength.

Also number of transitions are directly proportional to photons emitted.

So maximum of 3 photons gets emitted in total transition of electron from n=4 to 1.

Hope this helps.

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