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Question
a) What is the wavelength of the photon needed to ionize a hydrogen atom from its n=3 state?
b) The Balmer series for the hydrogen atom corresponds to transitions that terminate in the state with quantum number n=2.
i. What is the longest wavelength of light emitted in these transitions?
ii. What is the shortest wavelength of visible light emitted in these transitions? (note: visible light has wavelengths in the range of 390nm-700nm)

Thanks!

Answer
a) In order to ionize (remove) an electron from a hydrogen atom the incoming photon must have sufficient energy to remove the electron. An electron in the n=3 energy state has an energy:
U3=-13.6/n^2=-13.6/3^2=-1.51eV
Which is:
U3=-1.51eV*1.6x10^-19J/eV=2.4x10-19J
The wavelength of the required light photon will be:
h*c/lambda=U3
Therefore:, the required wavelength will be:
lambda=h*c/U3=6.63x10^-34*cx10^8/2.4x10-19=8.29x10-7m=829nm

For the Balmer spectrum do the same as above but where the minimum transition is between n=3 and n=2 for the red light and n=infinity and n=2 for the shortest wavelength.

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James J. Kovalcin

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I am teaching or have taught AP physics B and C [calculus based mechanics & electricity and magnetism] as well as Lab Physics for college bound students. I have a BS in Physics from the University of Pittsburgh and a Master of Arts in Teaching from same. I have been teaching physics for 34 years. I am constantly updating my skills and have a particular interest in modern physics topics.

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