Question

Part A A hydrogen atom makes a transition from the n = 5 state to a lower energy state. If the emitted photon has an energy of E = 0.306 eV, what is the quantum number of the lower state? n = 4 Submit Previous Answers Correct VP 28.6.2. Part B A hydrogen atom is excited into the n = 5 state and then decays to a lower energy state. What are the possible wavelengths of the emitted photon? Enter your answers in nanometers to three significant figures. Enter your answers in ascending order separated by commas. 11, 12, 13, 14 = 95.0,434,1280,4050 nm Submit Previous Answers

Answer 1

Part A I n 5 E= 0.306oV E = 13.6 ( 1 ) 0.306 = 13.6 ( 16 16 Anwer zfmf - Ķ ) alberov - Part B Posible waivelerighe. nia 5 For noi dillum na 2 'For 097.x107455 di 304 nm. for n=3 I 1.097X107/ 13 A3 = 684 nm

for nah . 1:097X10 1 5 - du- 1823 nma _ Anzuge : (14nm, 304nmub Blenm, 1823.nm) part c n = 4 _Tonixtion_energy E = me4 880²4² - hez 2 . hi E = 13.6 ( 4 ) E = 13 6 2 3.40V Answer & (3.4 ev)

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