A) \[\frac{25}{9}\]
B) \[\frac{17}{6}\]
C) \[\frac{9}{5}\]
D) \[\frac{4}{3}\]
Correct Answer: A
Solution :
\[\frac{1}{{{\lambda }_{\max }}}=R\left[ \frac{1}{{{4}^{2}}}-\frac{1}{{{5}^{2}}} \right]=\frac{9}{25\times 16}R\] and\[\frac{1}{{{\lambda }_{\min }}}=R\left[ \frac{1}{{{4}^{2}}}-\frac{1}{{{\infty }^{2}}} \right]=\frac{R}{16}\] \[\therefore \] \[\frac{{{\lambda }_{\max }}}{{{\lambda }_{\min }}}=\frac{25}{9}\]
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