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Hydrogen (""_(1)H^(1)), Deuterium (""_(1)H^(2)), single ionised Helium (""_(2)He^(4))^(+) and doubly ionised lithium (""_(3)Li^(6))^(++) all have one electron around the nucleus. Consider an electron transition from n=2 to n=1. If the wavelength of emitted radiation are lamda_(1),lamda_(2),lamda_(3) and lamda_(4) respectively then approximately which one of the following is corrent ? |
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Answer» `lamda_(1)=lamda_(2)=2lamda_(3)=lamda_(4)` `(1)/(lamda_(ik))=RZ^(2)[(1)/(n_(k)2)-(1)/(n_(i)2)]` here R and `n_(k),n_(i)` are constant, `:.lamda_(ik)prop(1)/(Z^(2))` `:.lamda_(1):lamda_(2):lamda_(3):lamda_(4)=(1)/((1)^(2)):(1)/((1)^(2)):(1)/((2)^(2)):(1)/((3)^(2))` `=(1)/(1):(1)/(1):(1)/(4):(1)/(9)` `:.lamda_(1):lamda_(2)=1:1` `:.lamda_(1)=lamda_(2)` `lamda_(1):lamda_(3)=1:(1)/(4)` `:.(lamda_(1))/(lamda_(3))=(1)/((1)/(4))implieslamda_(1)=4lamda_(3)` and `lamda_(1):lamda_(4)=1:(1)/((1)/(9))` `:.(lamda_(1))/(lamda_(4))=(1)/((1)/(9))implieslamda_(1)=9lamda_(4)` `:.lamda_(1)=lamda_(2)=4lamda_(3)=9lamda_(4)` |
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