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Author Question: The ionization energy for the hydrogen atom is 13.6 eV. What is the energy of a photon that is ... (Read 105 times)

mwit1967

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on: Jul 28, 2018
The ionization energy for the hydrogen atom is 13.6 eV. What is the energy of a photon that is emitted as a hydrogen atom makes a transition between the n = 4 and n = 3 states?
 a. 3.40 eV
  c. 6.80 eV
  b. 0.66 eV
  d. 2.55 eV

Question 2

The Lyman series of hydrogen is made up of those transitions made from higher levels to n = 1 . If the first line (n = 2 to n = 1) in this series has a wavelength of 122 nm, what is the wavelength of the third line (n = 4 to n = 1)?
 a. 364 nm
  c. 103 nm
  b. 97.6 nm
  d. 486 nm


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Juro

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Reply #1 on: Jul 28, 2018
Answer to Question 1

B

Answer to Question 2

B


mwit1967

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Reply 2 on: Jul 28, 2018
Wow, this really help

bigcheese9

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Reply 3 on: Yesterday
YES! Correct, THANKS for helping me on my review

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