the wavelength of the first line of lyman series is `1215 Å`, the wavelength of first line of balmer series will be
the wavelength of the first line of lyman series is `1215 Å`, the wavelength of first line of balmer series will be
A. `4545 Å`
B. `5295 Å`
C. `6561 Å`
D. `6750 Å`
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Correct Answer - C
Here`lambda_L=1215Å`
for the first line of lyman series.
`(1)/(lambda_L)=R[1/(1^2)-1/(2^2)]=R[1-1/(4)]=(3R)/(4)`
`therefore lambda_L=(4)/(3R)` for first line of balmer series
`(1)/(lambda_B)=R[1/(2^2)-1/(3^2)]=R[(1)/(4)-1/(9)]rArr(1)/(lambda_B)=R[5/(36)]`
`therefore lambda_B=(36)/(5R)`
From (i) and (ii), `(lambda_B)/(lambda_L)=(36//5R)/(4//3R)=(36xx3)/(4xx5)`
`therefore lambda_B=(108)/(20)xxlambda_L=(108)/(20)xx1215=6561Å`
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