In a hydrogen atom, an electron jumps from third orbit to the first orbit. Find out the frequency and wavelength of the spectral line. Given R=1.097xx 10^(7) m^(-1)

In a hydrogen atom, an electron jumps from third orbit to the first or

1.	In a hydrogen atom, an electron jumps from third orbit to the first orbit. Find out the frequency and wavelength of the spectral line. Given R=1.097xx 10^(7) m^(-1)
Answer» Solution :`1/lamda=barv=R[1/(n_(1)^(2))-1/(n_(2)^(2))]` `n_(1)=1,n_(2)=3` `1/lamda=1.097xx10^(7)m^(-1)[1-1/9]=0.9749xx10^(7)m^(-1)` `=1/(0.9747xx10^(7)m^(-1))=1.0254xx10^(-7)m=1025.4xx10^(-10)m=1025.4` Å `v=c/lamda=(3.0xx10^(8)ms^(-1))/(1.0254xx10^(-7)m)=2.9257xx10^(15)s^(-1)` Wavelength of the SPECTRAL line = 1025.4 Å Frequency of the spectral line `=2.9257xx10^(15)s^(-1)`