The lagest and shortest wavelengths in the Lyman series for hydrogen

Options

A. 1215 Å and 911 Å
B. 1315 Å and 900 Å
C. 1115 Å and 800 Å
D. 1015 Å and 850 Å

Answer

A

Explanation

The transition equation for Lyman series is given by,
`(1)/(lamda)=R((1)/(1^(2))-(1)/(n^(2))),n=2,3` . . .
The largest wavelength is corresponding to `n=2`.
`therefore (1)/(lamda_(max))=1.097xx10^(7)((1)/(1)-(1)/(4))=0.823xx10^(7)`m
`therefore lamda_(max)=1.2154xx10^(-7)m=1215` Å
The shortest wavelength corresponds to `n=oo`,
`therefore (1)/(lamda_("min"))=1.097xx10^(7)((1)/(1)-(1)/(oo))`
or `lamda_("min")=0.911x10^(-7)m=911` Å
Both of these wavelengths lie in ultraviolet (UV) region of electromagnetic spectrum.