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The electron energy in hydrogen atom is ...

The electron energy in hydrogen atom is given by `E_(n)=(-2.18xx10^(-18))//n^(2)J`. Calculate the energy required to remove an electron completely from the `n=2` orbit. What is the longest wavelength of light in cm that can be used to cause this transition?

Text Solution

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`E=-(21.7xx10^(-12))/(n^2)erg`
Electron energy in the 2nd orbit, i.e., n=2,
`E_(2)=-(21.7xx10^(-12))/(2^(2))erg=-5.425xx10^(-12)erg`
and `E_(infty)=0`
`DeltaE=` Change in energy `=E_(infty)-E_(2)=5.425xx10^(-12)erg`
Thus, energy required to remove an electron from 2 nd orbit
`=5.425xx10^(-12)erg`
According to quantum equation,
`DeltaE=h.(c)/(lamda)`
or `lamda=(hv)/(DeltaE)`
`(h=6.625xx10^(-27)erg-sec,c=3xx10^(10)cm//sec)`
and `DeltaE=5.425xx10^(-12)erg`
So, `lamda=((6.625xx10^(-27))xx(3xx10^(10)))/(5.425xx10^(-12))`
`=3.7xx10^(-5)cm`
Thus, the longest wavelength of light that can cause this transition is `3.7xx10^(-5)cm`.
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