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. (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? (Be sure to consider all values of n, not just those indicated by the arrows in the figure.) nm

Answer 1

Electronic transitions that end up in the state with quantum number n = 2 are what the Balmer series for the hydrogen atom refers to. The longest wavelength is 656.3x10-9 m, and the energy is  3.03X10^33J

Given quantum number (n) = 2

This will be transition 3-2 photon energy (E)= hc/λ since we must take into account the photon with the longest wavelength that completes the electron transition in the state with the quantum number n=2

photon energy (E)= hc/λ

(a) Then the energy (El) of longest wavelength= 6.63×10 −34 x 3x10^8/656.3x10-9

E = 3.03x10^33J = 1.89eV

(b) We know that 1/λl = R(1/n1^2 - 1/n2^2) where R =  Rydberg constant

1/λl = 1.097×107m−1× (1/2^2 - 1/3^2)= 656.3x10-9m

(c) From given diagram we say thee shortest transition is from level 6 to level 2

n1 = 2 and n2 = 6

Its photon energy (Es) =  6.63×10 −34 x 3x10^8/411x10-9 = 3.023eV

(d) Wavelength of shortest transition series 1/λ = 1.097×107x(1/2^2 -1/6^2) =  411nm

(e)Shortest wavelength is emitted in Balmer series if the transition of electron takes place from n2​=∞ to n1​=2.

Shortest wavelength in Balmer series 1/λ = R(1/2^2 -1/∞)=364.6nm

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