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If the electron jumps to the ground state from the third excited state in hydrogen atom, calculate the wavelength of corresponding emitted photon in eV. Rydberg's constant =1.097xx210^(7)m^(-1), c=3xx10^(8)m s^(-1)=6.62xx10^(-34)Js |
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Answer» Solution :`R=1.097xx10^(7)m^(-1)` `c=3xx10^(8)m s^(-1)` `h=6.62xx10^(-34)Js` `n_(i)=4` [ Third excited state =3+1] `n_(f)=1` `(1)/(lambda)=R[(1)/(n_(f)^(2))-(1)/(n_(i)^(2))]` `=1.097xx10^(7)[(1)/((1)^(2))-(1)/((4)^(2))]` `=1.097xx10^(7)[(15)/(16)]` `lambda=(16xx10^(-7))/(1.097xx15)` `lambda=9.72xx10^(-8)m` `E=(HC)/(lambda)=(6.62xx10^(-34)xxx3xx10^(8))/(9.72xx10^(-8))` `=2.04xx10^(-18)J` `=(2.04xx10^(-18))/(1.6xx10^(-19))eV` `:.E=1.277eV`s |
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