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After absorbing an energy of 20.44xx10^(...

After absorbing an energy of `20.44xx10^(-19)J`, the electron of H-atom will jump to which orbit ?

Text Solution

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The energy of the electron in the n-th orbit of H-atom is,
`E_(n)=(-2.18xx10^(-19))/(n^(2))J`
The energy of an electron in the ground state (n=1) is,
`E_(1)=-21.8xx10^(-19)J`
If the electron absorbs an energy of `2..044xx10^(-19)J`, the total energy `=(-21.8xx10^(-19)+20.44xx10^(-19))J=-1.36xx10^(-19)J`
`therefore (-21.8xx10^(-19))/(n^(2))=-1.36xx10^(-19)J`
`therefore n=sqrt((-21.8xx10^(-19))/(-1.36xx10^(-19)))=4`
thus, the electron will jump to the fourth orbit.
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