6:21 PM iPad 6% webassign net Need Help? Read It 2.34/5.88 points l Previous Ans
ID: 1393240 • Letter: 6
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6:21 PM iPad 6% webassign net Need Help? Read It 2.34/5.88 points l Previous Answers serpop5 29.p.006.nva My Notes Ask Your Teacher The Balmer series for the hydrogen atom corresponds to electronic transitions that terminate in the state with quantum number n 2 as shown in the figure below. Consider the photon of longest wavelength corresponding to a transition shown n the figure E (eV) 0.00 0.544 2 0.850 4 4 1.512 3.401 Balmer Series (a) Determine its energy eV 1.889 (b) Determine its wavelength. 656.35 nm Consider the spectra line of shortest wavelength corresponding to a transition shown in the figure. (Be sure to consider all values of n, not just those indicated by the arrows in the figure.) (c) Find its photon energy 2.8568 Your response differs from the correct answer by more than 10%. Double check your calculations. eV (d) Find its wavelength. 434 Your response differs from the correct answer by more than 10%. Double check your calculations. nm (e) What is the shortest possible wavelength in the Balmer series? nm +-5.88 points SerPOP5 29.F.007 My Notes Ask Your Teacher A photon with energy 2.90 eV is absorbed by a hydrogen atom. (a) Find the minimum n for a hydrogen atom that can be ionized by such a photon.Explanation / Answer
c) For shortest wavelength enrgy should be maximum.
When electron moves from n=infinte state to n = 2,
it emits with maximum enrgy
so,
Energy of phton = 3.401 eV <<<<<-----Answer
d)
E = h*c/lamda
==> lamda = h*c/E
= 6.625*10^-34*3*10^8/(3.401*1.6*10^-19)
= 3.652*10^-7 m or 365.2 nm <<<<<-----Answer
e) 365.2 nm <<<<<-----Answer
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