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This is from Advanced Physics by Adams and Allday, part 8 Modern Physics, Practice Exam Question 15.

- The spectrum of atomic hydrogen contains a prominent red line having a wavelength of 6.60 x 10
^{-7}m. Calculate the energy of a photon with this wavelength.- The ionisation energy of hydrogen is 2.18 x 10
^{-18}J. The next allowed energy level above the ground state in hydrogen has an energy -5.40 x 10^{-19}J. Show by calculation that the lowest energy level cannot be involved in the production of the prominent red line in a.

2. Relevant equations

E = h f

v = f λ

3. The attempt at a solution

f = v / λ

E = h v / λ

= 6.63E-34 * 3.00E+8 / 6.60E-7

= 3.0E-19 J ct2sf (Book gives same answer. Calculated 3.013636364e-19)

A drop from the next allowed energy level above the ground state to the ground state would release 5.40E-19 J. This does not match energy calculated in a so this drop cannot be the one that produces the red line having a wavelength of 6.60E-7 m.

A free electron dropping to the next allowed energy level above the ground state would release 2.18E-18 - 5.40E-19 = 1.64E-18 J. This is more than the energy that produces the red line having a wavelength of 6.60E-7 m so I cannot, on the available data, show that some drop down to the next allowed energy level above the ground state does not produce the red line having a wavelength of 6.60E-7.

Hmm ...

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# Homework Help: Hydrogen spectrum calculations

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