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Problem 4: The series limit wavelength of the Balmer series is emitted as the electron in the hydrogen atom falls from n = ∞ to the n = 2 state. What is the wavelength of this line where ΔE = 3.40 eV.

Solution

When know that when an electron falls from high energy level to a lower one, it emits a photon of energy equal to the difference of energies between the two states. In the given case, electron falls from an infinite level to level 2 and emits energy ΔE = 3.40 eV as radiation (in the Balmer series). Now,

However, we have to convert energy from electron volts to joules first. As 1 eV = 1.602 × 10-19 J ⇒ 3.40 eV = 3.40 × 1.602 × 10-19 J. Now put the values as h = 6.63 × 10-34 J s, and c = 3 × 108 ms-1.

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