# The ratio of minimum to maximum wavelengths of radiation that an electron cause in Bohr's hydrogen atom is

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

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 )
$\varepsilon_{max}=13.6\;(1-\large\frac{1}{\infty})=13.6eV$
$\large\frac{\varepsilon_{min}}{\varepsilon_{max}}=\large\frac{3}{4} \to \large\frac{\large\frac{hc}{\lambda_{max}}}{\large\frac{hc}{\lambda_{min}}}$
$\large\frac{\lambda_{min}}{\lambda_{max}}=\large\frac{3}{4}\;.$