question_answer

If the wavelength of 1st line of Balmer series is \[6563\,\overset{o}{\mathop{A}}\,\], the wavelength of first line of Lyman series and Rydbergs constant respectively, will be:

A)  \[1215.4\,\overset{o}{\mathop{A}}\,,\,1.1\times {{10}^{7}}{{m}^{-1}}\]

B)  \[5863\,\overset{o}{\mathop{A}}\,,\,2.0\times {{10}^{7}}{{m}^{-1}}\]

C)  \[2316.4\,\overset{o}{\mathop{A}}\,,\,1.0\times {{10}^{7}}{{m}^{-1}}\]

D)  none of the above

Correct Answer: A

Solution :

 The wavelength of spectral line of Balmer series is given by \[\frac{1}{{{\lambda }_{B}}}=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right)=\frac{5R}{36}\] ? (i) where R is Rydbergs constant. For Lyman series \[\frac{1}{{{\lambda }_{L}}}=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{2}^{2}}} \right)=\frac{3R}{4}\] ? (i) \[\therefore \] \[\frac{1/{{\lambda }_{B}}}{1/{{\lambda }_{L}}}=\frac{5R/36}{3R/4}=\frac{5}{27}\] \[\Rightarrow \] \[{{\lambda }_{L}}=\frac{5}{27}{{\lambda }_{B}}=\frac{5}{27}\times 6563=1215.4\,\overset{o}{\mathop{A}}\,\] Rydbergs constant \[R=\frac{36}{5\,{{\lambda }_{B}}}=\frac{36}{5\times 6563\times {{10}^{-10}}}\] \[=1.098\times {{10}^{7}}\,{{m}^{-1}}\] \[\approx 1.1\times {{10}^{7}}{{m}^{-1}}\]