A) 1215\[\overset{0}{\mathop{A}}\,\]
B) 4861\[\overset{0}{\mathop{A}}\,\]
C) 6050\[\overset{0}{\mathop{A}}\,\]
D) data given is insufficient to calculate the value
Correct Answer: B
Solution :
Balmer found that the wavelength of all lines in hydrogen spectrum can be represented by \[\frac{1}{\lambda }=R\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] where R is Rydbergs constant. Given, \[{{n}_{2}}=3,\,{{n}_{1}}=2\] \[\therefore \] \[\frac{1}{{{\lambda }_{1}}}=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right)=\frac{5R}{36}\] \[\Rightarrow \] \[\frac{1}{{{\lambda }_{2}}}=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right)=\frac{3R}{16}\] \[\Rightarrow \] \[\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}=\frac{16}{3R}\times \frac{5R}{36}\] \[\Rightarrow \] \[{{\lambda }_{2}}=\frac{80}{108}{{\lambda }_{1}}\] Given, \[{{\lambda }_{1}}=6563{\AA}\] \[\therefore \] \[{{\lambda }_{2}}=\frac{80}{180}\times 6563=4861{\AA}\]
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