The Balmer series for the hydrogen atom corresponds to electronic transitions th

Question

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 in the figure. E(eV) 0.00 6 0.378 0.5442 0.8504 1.512 3.401 Balmer Series (a) Determine its energy. eV (b) Determine its wavelength. nm Consider the spectral line of shortest wavelength corresponding to a transition shown in the figure. (C) Find its photon energy. eV (d) Find its wavelength. nm (e) What is the shortest possible wavelength in the Balmer series? nm

Explanation / Answer

if electron make a transition from n=3 to n= 2

we get longest wavelength

(a) En = -13.6/n^2

E3 - E2 = -13.6(1/3^2 -1/2^2)

= 1.89 eV

(b) DE = hc/lamda

1.89 = 1243/lamda

lama = 657.67 nm

(c) if electron make a transition from n=infinity to n= 2

we get shortest wavelength

(d) En = -13.6/n^2

DE = -13.6(0-1/2^2)

= 3.4 eV

(d) DE = hc/lamda

3.4 = 1243/lamda

lamda = 365.6 nm

(e) lamda = 365.6 nm

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