| Statement I: Balmer series lies in the visible region of electromagnetic spectrum. |
| Statement II: \[\frac{1}{\lambda }=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{n}^{2}}} \right).\]where n = 3,4,5. |
A) Statement I is true; Statement II is true; Statement II is not a correct explanation for Statement I.
B) Statement I is true; Statement II is false.
C) Statement I is false; Statement II is true.
D) Statement I is true; Statement II is true; Statement II is not a correct explanation for Statement I.
Correct Answer: A
Solution :
The wavelength in Balmer series is given by \[\frac{1}{\lambda }=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{n}^{2}}} \right),n=3,4,5,...\] \[\frac{1}{{{\lambda }_{\max }}}=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right)\] \[{{\lambda }_{\max }}=\frac{36}{5R}\] \[=\frac{36\times 1}{5\times 1.097\times {{10}^{7}}}=6563{\AA}\] and \[\frac{1}{{{\lambda }_{\min }}}=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{\infty }^{2}}} \right)\] \[{{\lambda }_{\min }}=\frac{4}{R}=\frac{4}{1.097\times {{10}^{7}}}=3646{\AA}\] The wavelength, 6563 Å and 3646 Å lie in visible region. Therefore, Balmer series lies in visible region.
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