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Calculate the De-broglie Wavelength Associated with the Electron Revolving in the First Excited State of Hydrogen Atom. the Ground State Energy of the Hydrogen Atom - Physics

Short Note

Answer the following question.
Calculate the de-Broglie wavelength associated with the electron revolving in the first excited state of the hydrogen atom. The ground state energy of the hydrogen atom is – 13.6 eV.

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Solution

de-Broglie wavelength `lambda = h/(mv) = h/p`, where p is momentum of electron
Kinetic energy (KE) and momentum (p) are related by, `"KE" = p^2/(2m) ("m"="mass")`
⇒ `p = sqrt(2m("KE")`
⇒ `lambda = h/sqrt(2m("KE")`
According to Bohr's model, `"Kinetic Energy of"  e^-  = |"Total Energy of " e^-  |=|-(13.6xxZ^2)/(n^2)|e"V"`
for Hydrogen Z = 1 and first excited state implies n = 2

`"KE"  = (13.6xx1^2)/(2^2) = 3.4  "eV"`
`= 3.4 xx 1.6 xx 10^-19 "J"`
`= 5.44 xx 10^-19 "J"`
putting the values  in formula for wavelength we get,
`lambda = h/sqrt(2m("KE"))=(6.63xx10^-34)/sqrt(2xx9.1xx10^-31xx5.44xx10^-19)=6.66xx10^-10"m" = 6.66Å`

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