Question

A Hydrogen sample is in 1st excited state. A photon of energy 8eV is used to excited the sample. Find the De-Broglie wavelength of the electron.​

Answer 1

Answer:

Here is the answer to your question:

electron in 1st excited state: n=2

energy in this state= -13.6/(n^2) eV

= -13.6/2^2 = 3.4eV

now energy provided = 8eV

extra energy left with electron after getting ionisation energy of +3.4eV is (8-3.4)eV = 4.6eV

energy= hc/(wavelength)

wavelength comes out to be : 2703 nm

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