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A hydrogen atom initially in the ground ...

A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength andfrequency of photon.

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Energy of electron in `n^(th)` orbit of atom
`E_(n) = - (13.6)/(n^(2))eV`
`rArr` The energy of electron in ground state
`E_(1)=-(13.6)/((1)^(2))eV [ :.` For ground state n=1]
`:.E_(1)=-13.6eV`
`rArr` Energy of electron in n=4 orbit,
`E_(4)=-(13.6)/((4)^(2))eV`
`=-(13.6)/(16)eV`
`:.E_(4)=-0.85ev`
`rArr` Different in enegy of electron,
`DeltaE=E_(4)-E_(1)`
`=-0.85eV-(13.eV)`
`=12.75 eV` br> `=12.75xx1.6xx10^(-19)J`
`=20.4xx10^(-19)J`
`rArr` Now,
`DeltaE=hv`
`:.v=(DeltaE)/(h)=(20.4xx10^(-19))/(6.625xx10^(-34))`
`:.v=3.073xx10^(15)Hz`
`rArr` , now
`c= lambda v`
`:. lambda (c)/(v)=(3xx10^(8))/(30.79xx10^(15))`
`:. lambda=97.43xx10^(-9)=97.43 nm`
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