Question

In a hydrogen atom, in transition of electron a photon of energy 2.55 ev is emitted, then the change in wavelength of the electron is

Answer 1

The change in wavelength is 56.0
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Answer 2

Answer:

The energy of the photon is ΔE=2.55eV=2.55×1.602×10−19J, ΔE=2.55eV=2.55×1.602×10−19J.

The wavelength of the photon of light is λ=hc/ΔE=(6.62×10−34×3×108)/(2.55×1.602×10−19)m=4861 A0  

This corresponds to the second line of Balmar series. The transition is n2=4→n1=2

The change in the wavelength of the electron is Δλ=h/mu4−h/mu2=h/m(1/u4−1/u2)=(6.626×10−34)× (4−2)/(9.1085×10−31  ×(2.19×108)=6.64A0