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The first member of the Balmer series of...

The first member of the Balmer series of hydrogen atom has wavelength of 656.3nm. Calculate the wavelength and frequency of the second member of the same series. Given, `c=3xx10^(8)m//s`.

Text Solution

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Given `lamda=656.3xx10^(-9)m`
`c=3xx10^(8)ms^(-1)`
For balner series
`n_(1)=2` First member `n_(2)=3`
w:k.t. `1/lamda=R((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))`
`1/(656.3xx10^(-9))=R((1)/(2^(2))-(1)/(3^(2)))`
`1/(656.3xx10^(-9))=R(5/56)`
`:. R=36/(5xx656.3xx10^(-9))`
`R=1.097xx10^(7)m^(-1)`
For second member `n_(2)=4`
`:.1/lamda=1.097xx10^(7)xx(1/2^(2)-1/4^(2))`
`=(1.097xx10^(7)xx12)/(64)`
`:.lamda=64/(12xx1.097xx10^(7))m=4.8617xx10^(7)m`
`v=C/lamda=(3xx10^(8))/(4.8617)xx10^(7)Hz=0.6171xx10^(15)Hz " or " v = 6.17xx10^(14)Hz`
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