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InterviewSolution
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The first member of the Balmer series of hydrogen atom has wavelength of 656.3 nm. Calculate the wavelength and frequency of the second member of the same series. Given, c=3xx10^(8)m//s. |
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Answer» Solution :`1/lambda_1=R(1/n_1^2-1/n_2^2)` For I member of Balmer series, `n_1=2` & `n_2=3, lambda_1` = 6563 Å `1/lambda_1=R(1/4-1/9)=(5R)/36` For II member of Balmer series, `n_1`=2 and `n_2`=4 `1/lambda_1=R(1/4-1/16)=(3R)/16` Arriving at `lambda_2`= 4861Å `v=c//lambda_2=6.1xx10^14` Hz Given `lambda_0`=6563 Å=`6.536xx10^(-7)` m `lambda_beta` =? `f_beta` = ? `c=3xx10^(8) ms^(-1)` `rArr 1/lambda_1 =R(1/n_1^2-1/n_2^2)` For I member of Balmer series, `n_1=2,n_2=3` `therefore 1/(6xx563xx10^(-7))=R(1/2^2-1/3^2)` i.e., `10^7/6.563=R(5/36)` HENCE `R=36/5xx10^7/6.563` i.e., `R=1.097xx10^7 m^(-1)` For II member of Balmer series `x_1=2, x_2=4` `therefore 1/lambda_2=1.097xx10^7(1/2^2-1/4^2)` i.e., `1/lambda_2=(1.097xx10^7xx12)/64` Hence `lambda_2=64/(1.097xx12xx10^7)` m i.e., `lambda_2=4.8617xx10^(-7)` m `lambda_2`=4861.7Å and frequency , `Y_2=c/lambda_2=(3xx10^8)/(4.8617xx10^(-7))` i.e., `Y_2=0.6171xx10^15` Hz = `6.171xx10^14` Hz |
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