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In the line spectra of hydrogen atom, di...

In the line spectra of hydrogen atom, difference between the largest and the shortest wavelengths of the Lyman series is `340Å` . The corresponding difference of the Paschan series in `Å` is ...........

Text Solution

Verified by Experts

The correct Answer is:
`10553.14`
`lambda =c/((1/(n_(1)^(2))-1/(n_(2)^(2))))`
For Lyman series : `lambda_(1)=c/(1/(1^(2)) -1/(oo^(2))) =c " " ( n oo " to "n=1)`
`lambda_(2)=c/(1/1^(2)-1/(2^(2)))=(4c)/3 " " (n = 2 " to "n=1)`
`Deltalambda = lambda_(2) -lambda_(1) = c/3 =304 Å`
`rArr c = 912 Å`
For Paschen series :
`lambda_(1)=c/(1/(3^(2))-1/(4^(2)))=(144c)/7 " " (n = 4 " to " n = 3)`
`Deltalambda = lambda_(2) -lambda_(1) = (144c)/7 - 9c =(81c)/7 = (81xx912)/7`
`= 10553.14 Å`
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