$(a)\;\large\frac{1}{2}\qquad(b)\;zero\qquad(c)\;\large\frac{3}{4}\qquad(d)\;\large\frac{27}{32}$

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Answer : (c) $\;\large\frac{3}{4}$

Explanation :

Energy of realiation that corresponds to energy difference between two energy levels $\;n_{1}\;$ and $\;n_{2}\;$ is given as

$\varepsilon=13.6\;(\large\frac{1}{n_{1}^2}-\large\frac{1}{n_{2}^2}) eV$

$\varepsilon \;$ is minimum when $\;n_{1}=1\;$ & $\;n_{2}=2$

$\varepsilon_{min}=13.6\;(1-\large\frac{1}{4})eV$

$=13.6\times \large\frac{3}{4} eV$

$\varepsilon\;$ is maximum when $\;n_{1}=1\;$ & $\;n_{2}=\infty\;$ (the atom is ionized , thatt is known as ionization energy )

$\varepsilon_{max}=13.6\;(1-\large\frac{1}{\infty})=13.6eV$

$\large\frac{\varepsilon_{min}}{\varepsilon_{max}}=\large\frac{3}{4} \to \large\frac{\large\frac{hc}{\lambda_{max}}}{\large\frac{hc}{\lambda_{min}}}$

$\large\frac{\lambda_{min}}{\lambda_{max}}=\large\frac{3}{4}\;.$

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