A) \[6557\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
B) \[1216\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
C) \[4800\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
D) \[5600\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
Correct Answer: A
Solution :
The longest wavelength in Balmer series of hydrogen spectrum occurs in third orbit. The relation for Balmer series is given by \[\frac{1}{\lambda }=R\left[ \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right]\] \[=R\left[ \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{3}}} \right]\] (where R has the value \[1.098\times {{10}^{7}}\] per metre) \[=\frac{5}{36}\times 1.098\times {{10}^{7}}\] \[\lambda =\frac{36\times {{10}^{-7}}}{5\times 1.098}\] \[=6557\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
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