1.

What is the wavelength of light for the least energetic photoa emitted in the Lyman series of the hydrogen spectrum. (take hc = 1240 eV nm):

Answer»

102nm
150nm
82nm
122nm

Solution :For least ENERGETIC PHOTON in Lyman series
`n_(1)=1 and n_(2)=2`
then `E_(1)=-13.6eV and E_(2)=-3.4eV`
`E=E_(2)-E_(1)=-3.4 +13.6=10.2eV`
Now `E=hv=(HC)/(lambda)`
`lambda=(hc)/(E)=(1240eV)/(10.2eV)NM`
`lambda=122nm`.


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