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The ground state energy of hydrogen atom...

The ground state energy of hydrogen atom is -13.6eV. If an electron makes a transition form an energy level -0.85 eV to -3.4 eV, calculate the wavelength of spectral line emitted. To which series of hydrogen spectrum does this wavelength belongs?

Text Solution

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Here, `E=-13.6eV=(-13.6)/(n^2)eV`,
Where n=1
Now, `=-0.85eV=(-13.6)/(n_1^2) :. n_1^2=13.6/0.85=16`,
and `-3.4eV =(-13.6)/(n_2^2) :. n_2^2=13.6/3.4=4,`
`n_2=2`
In transition form `n_1` to `n_2`,
`hv=(hc)/lambda=E_1-E_2=-0.85-(-3.4)`
`=2.55eV=2.55xx1.6xx10^(-19)J`
`lambda=(hc)/(2.55xx1.6xx10^(-19))=(6.6xx10^(-34)xx3xx10^8)/(2.55xx1.6xx10^(-19))`
`=4.852xx10^-7m =4852Å`
This wavelength belongs to visible region of hydrogen spectrum.
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